Skip to main content

Math Finite Mathematics (11th Edition)

(a) To set up: The linear programming problem to maximize profit. Solution: Maximize x 1 + 1.5 x 2 = z _ Subjected to x 1 + 2 x 2 ≤ 200 4 x 1 + 3 x 2 ≤ 600 0 ≤ x 2 ≤ 90 x 1 ≥ 0 _ Given: A small toy manufacturing firm has 200 squares of felt, 600 oz of stuffing, and 90ft of trim. The small bear required 1 square of felt and 4 oz of stuffing. The monkey required 2 squares of felt, 3 oz of stuffing, and 1 ft of trim. Profit on each bear is $1 and profit on each monkey is $1.5. Explanation: Let x 1 = a small bear x 2 = a monkey Maximize x 1 + 1.5 x 2 = z Subjected to x 1 + 2 x 2 ≤ 200 4 x 1 + 3 x 2 ≤ 600 0 ≤ x 2 ≤ 90 x 1 ≥ 0 Conclusion: Above in the explanation, write set up the linear programming problem to maximize. (b) To solve: The linear programming problem in part (a). Solution: The maximum value of the objective function is 180 at the corner point ( 120 , 40 ) _ . Given: Provided in part (a) Maximize x 1 + 1.5 x 2 = z Subjected to x 1 + 2 x 2 ≤ 200 4 x 1 + 3 x 2 ≤ 600 0 ≤ x 2 ≤ 90 x 1 ≥ 0 Explanation: Provided in part (a) The constraints represented in the form of linear inequalities are Maximize x 1 + 1.5 x 2 = z Subjected to x 1 + 2 x 2 ≤ 200 4 x 1 + 3 x 2 ≤ 600 0 ≤ x 2 ≤ 90 x 1 ≥ 0 The linear inequalities are plotted on the graph: The feasible region is the bounded region with corner points: ( 0 , 90 ) , ( 20 , 90 ) , ( 120 , 40 ) , ( 150 , 0 ) _ . The values of the objective function at the corner points are represented in the table: Corner points Values of objective function x 1 + 1.5 x 2 = z ( 0 , 90 ) _ 0 + 1.5 × 90 = 135 ( 20 , 90 ) _ 20 + 1.5 × 90 = 155 ( 120 , 40 ) _ 120 + 1.5 × 40 = 180 ( 150 , 0 ) _ 150 + 1.5 × 0 = 150 The maximum value of the objective function is 180 at the corner point ( 120 , 40 ) _ . Conclusion: Hence, themaximum value of the objective function is 180 at the corner point ( 120 , 40 ) _ . (c) To determine: The corresponding dual problem Solution: Minimize w =200 y 1 + 600 y 2 + 90 y 3 _ Subjected to y 1 + 4 y 2 ≥ 1 2 y 1 + 3 y 2 + y 3 ≤ 1.5 y 1 ≥ 0 , y 2 ≥ 0 , y 2 ≥ 0 _ Given: A small toy manufacturing firm has 200 squares of felt, 600 oz of stuffing, and 90 ft of trim. The small bear required 1 square of felt and 4 oz of stuffing. The monkey required 2 squares of felt, 3 oz of stuffing, and 1 ft of trim. Profit on each bear is $1 and profit on each monkey is $1.5. Explanation: The corresponding dual problem is as follows: Minimize w =200 y 1 + 600 y 2 + 90 y 3 Subjected to y 1 + 4 y 2 ≥ 1 2 y 1 + 3 y 2 + y 3 ≤ 1.5 y 1 ≥ 0 , y 2 ≥ 0 , y 3 ≥ 0 Conclusion: Above in the explanation, write the corresponding dual problem. (d) To determine: The optimal solution to the dual problem. Solution: The minimum value of the objective function is 180 at the corner point ( 0.6 , 0.1 , 0 ) _ . Given: Provided in part (c): Minimize w =200 y 1 + 600 y 2 + 90 y 3 Subjected to y 1 + 4 y 2 ≥ 1 2 y 1 + 3 y 2 + y 3 ≤ 1.5 y 1 ≥ 0 , y 2 ≥ 0 , y 3 ≥ 0 Explanation: Provided in part (c): The constraints represented in the form of linear inequalities are as follows: Minimize w =200 y 1 + 600 y 2 + 90 y 3 Subjected to y 1 + 4 y 2 ≥ 1 2 y 1 + 3 y 2 + y 3 ≤ 1.5 y 1 ≥ 0 , y 2 ≥ 0 , y 3 ≥ 0 The linear inequalities are plotted on the graph: The feasible region is the bounded region with corner points: ( 0 , 0 .5 , 0 ) , ( 0 , 6 , 0.1 , 0 ) , ( 1 , 0 , 0 ) . The values of the objective function at the corner points are represented in the table: Corner points Values of objective function w =200 y 1 + 600 y 2 + 90 y 3 ( 0 , 0.5 , 0 ) _ 200 × 0 + 600 × 0.5 + 90 × 0 = 300 ( 0.6 , 0.1 , 0 ) _ 200 × 0.6 + 600 × 0.1 + 90 × 0 = 180 ( 1 , 0 , 0 ) _ 200 × 1 + 600 × 0 + 90 × 0 = 200 The minimum value of the objective function is 180 at the corner point ( 0.6 , 0.1 , 0 ) _ . Conclusion: Hence, the minimum value of the objective function is 180 at the corner point ( 0.6 , 0.1 , 0 ) _ . (e) To determine: The profit the firm will make if its supply of felt increases to 210 squares by using shadow profits. Solution: The profit is $ 186 _ . Given: If its supply of felt increases to 210 squares. Explanation: If the firm will make its supply of felt increases to 210 squares. The value for felt is 0.6. An increase in the supply of 10 units of felt will increase profit to $ 180 + $ ( 0.6 × 10 ) = $ 180 + $ 6 = $ 186 Conclusion: Hence, the profit is $ 186 _ . (f) To determine: How much profit will the firm make if its supply of stuffing is cut to 590 oz and its supply of trim is cut to 210 squares. Solution: The profit is $ 179 _ . Given: If its supply of stuffing is cut to 590 oz and its supply of trim is cut to 210 squares. Explanation: If its supply of stuffing is cut to 590 oz and its supply of trim is cut to 210 squares. The shadow value is 0.1 for stuffing and 0 for trim. If stuffing and trim are each decreased by 10 units, the profit will be $ 180 − $ ( 0.1 × 10 ) − $ ( 0 × 10 ) = $ 180 − $ 1 = $ 179 Conclusion: Hence, the profit is $ 179 _ . (g) To explain: Why it makes sense that the shadow profit for trim is 0. Explanation: In this, y 3 = trim . From the above final tableau, the optimal solution to the dual problem is y 3 = 0 . Therefore, shadow profit trim is 0.

(a) To set up: The linear programming problem to maximize profit. Solution: Maximize x 1 + 1.5 x 2 = z _ Subjected to x 1 + 2 x 2 ≤ 200 4 x 1 + 3 x 2 ≤ 600 0 ≤ x 2 ≤ 90 x 1 ≥ 0 _ Given: A small toy manufacturing firm has 200 squares of felt, 600 oz of stuffing, and 90ft of trim. The small bear required 1 square of felt and 4 oz of stuffing. The monkey required 2 squares of felt, 3 oz of stuffing, and 1 ft of trim. Profit on each bear is $1 and profit on each monkey is $1.5. Explanation: Let x 1 = a small bear x 2 = a monkey Maximize x 1 + 1.5 x 2 = z Subjected to x 1 + 2 x 2 ≤ 200 4 x 1 + 3 x 2 ≤ 600 0 ≤ x 2 ≤ 90 x 1 ≥ 0 Conclusion: Above in the explanation, write set up the linear programming problem to maximize. (b) To solve: The linear programming problem in part (a). Solution: The maximum value of the objective function is 180 at the corner point ( 120 , 40 ) _ . Given: Provided in part (a) Maximize x 1 + 1.5 x 2 = z Subjected to x 1 + 2 x 2 ≤ 200 4 x 1 + 3 x 2 ≤ 600 0 ≤ x 2 ≤ 90 x 1 ≥ 0 Explanation: Provided in part (a) The constraints represented in the form of linear inequalities are Maximize x 1 + 1.5 x 2 = z Subjected to x 1 + 2 x 2 ≤ 200 4 x 1 + 3 x 2 ≤ 600 0 ≤ x 2 ≤ 90 x 1 ≥ 0 The linear inequalities are plotted on the graph: The feasible region is the bounded region with corner points: ( 0 , 90 ) , ( 20 , 90 ) , ( 120 , 40 ) , ( 150 , 0 ) _ . The values of the objective function at the corner points are represented in the table: Corner points Values of objective function x 1 + 1.5 x 2 = z ( 0 , 90 ) _ 0 + 1.5 × 90 = 135 ( 20 , 90 ) _ 20 + 1.5 × 90 = 155 ( 120 , 40 ) _ 120 + 1.5 × 40 = 180 ( 150 , 0 ) _ 150 + 1.5 × 0 = 150 The maximum value of the objective function is 180 at the corner point ( 120 , 40 ) _ . Conclusion: Hence, themaximum value of the objective function is 180 at the corner point ( 120 , 40 ) _ . (c) To determine: The corresponding dual problem Solution: Minimize w =200 y 1 + 600 y 2 + 90 y 3 _ Subjected to y 1 + 4 y 2 ≥ 1 2 y 1 + 3 y 2 + y 3 ≤ 1.5 y 1 ≥ 0 , y 2 ≥ 0 , y 2 ≥ 0 _ Given: A small toy manufacturing firm has 200 squares of felt, 600 oz of stuffing, and 90 ft of trim. The small bear required 1 square of felt and 4 oz of stuffing. The monkey required 2 squares of felt, 3 oz of stuffing, and 1 ft of trim. Profit on each bear is $1 and profit on each monkey is $1.5. Explanation: The corresponding dual problem is as follows: Minimize w =200 y 1 + 600 y 2 + 90 y 3 Subjected to y 1 + 4 y 2 ≥ 1 2 y 1 + 3 y 2 + y 3 ≤ 1.5 y 1 ≥ 0 , y 2 ≥ 0 , y 3 ≥ 0 Conclusion: Above in the explanation, write the corresponding dual problem. (d) To determine: The optimal solution to the dual problem. Solution: The minimum value of the objective function is 180 at the corner point ( 0.6 , 0.1 , 0 ) _ . Given: Provided in part (c): Minimize w =200 y 1 + 600 y 2 + 90 y 3 Subjected to y 1 + 4 y 2 ≥ 1 2 y 1 + 3 y 2 + y 3 ≤ 1.5 y 1 ≥ 0 , y 2 ≥ 0 , y 3 ≥ 0 Explanation: Provided in part (c): The constraints represented in the form of linear inequalities are as follows: Minimize w =200 y 1 + 600 y 2 + 90 y 3 Subjected to y 1 + 4 y 2 ≥ 1 2 y 1 + 3 y 2 + y 3 ≤ 1.5 y 1 ≥ 0 , y 2 ≥ 0 , y 3 ≥ 0 The linear inequalities are plotted on the graph: The feasible region is the bounded region with corner points: ( 0 , 0 .5 , 0 ) , ( 0 , 6 , 0.1 , 0 ) , ( 1 , 0 , 0 ) . The values of the objective function at the corner points are represented in the table: Corner points Values of objective function w =200 y 1 + 600 y 2 + 90 y 3 ( 0 , 0.5 , 0 ) _ 200 × 0 + 600 × 0.5 + 90 × 0 = 300 ( 0.6 , 0.1 , 0 ) _ 200 × 0.6 + 600 × 0.1 + 90 × 0 = 180 ( 1 , 0 , 0 ) _ 200 × 1 + 600 × 0 + 90 × 0 = 200 The minimum value of the objective function is 180 at the corner point ( 0.6 , 0.1 , 0 ) _ . Conclusion: Hence, the minimum value of the objective function is 180 at the corner point ( 0.6 , 0.1 , 0 ) _ . (e) To determine: The profit the firm will make if its supply of felt increases to 210 squares by using shadow profits. Solution: The profit is $ 186 _ . Given: If its supply of felt increases to 210 squares. Explanation: If the firm will make its supply of felt increases to 210 squares. The value for felt is 0.6. An increase in the supply of 10 units of felt will increase profit to $ 180 + $ ( 0.6 × 10 ) = $ 180 + $ 6 = $ 186 Conclusion: Hence, the profit is $ 186 _ . (f) To determine: How much profit will the firm make if its supply of stuffing is cut to 590 oz and its supply of trim is cut to 210 squares. Solution: The profit is $ 179 _ . Given: If its supply of stuffing is cut to 590 oz and its supply of trim is cut to 210 squares. Explanation: If its supply of stuffing is cut to 590 oz and its supply of trim is cut to 210 squares. The shadow value is 0.1 for stuffing and 0 for trim. If stuffing and trim are each decreased by 10 units, the profit will be $ 180 − $ ( 0.1 × 10 ) − $ ( 0 × 10 ) = $ 180 − $ 1 = $ 179 Conclusion: Hence, the profit is $ 179 _ . (g) To explain: Why it makes sense that the shadow profit for trim is 0. Explanation: In this, y 3 = trim . From the above final tableau, the optimal solution to the dual problem is y 3 = 0 . Therefore, shadow profit trim is 0.

Solution Summary: The author explains how to set up the linear programming problem to maximize profit.

Finite Mathematics (11th Edition)

Finite Mathematics (11th Edition)

11th Edition

ISBN: 9780321979438

Author: Margaret L. Lial, Raymond N. Greenwell, Nathan P. Ritchey

Publisher: PEARSON

See similar textbooks

Concept explainers

The word “unitary” comes from the word “unit”, which means a single and complete entity. In this method, we find the value of a unit product from the given number of products, and then we solve for the other number of products.

Speed, Time, and Distance

Imagine you and 3 of your friends are planning to go to the playground at 6 in the evening. Your house is one mile away from the playground and one of your friends named Jim must start at 5 pm to reach the playground by walk. The other two friends are 3 …

Profit and Loss

The amount earned or lost on the sale of one or more items is referred to as the profit or loss on that item.

Units and Measurements

Measurements and comparisons are the foundation of science and engineering. We, therefore, need rules that tell us how things are measured and compared. For these measurements and comparisons, we perform certain experiments, and we will need the experime…

Videos

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

Blurred answer

Chapter 4.3, Problem 21E

Chapter 4.3, Problem 23E

Students have asked these similar questions

Prove that a relation X defined on a set A that is reflexive, symmetric and antisymmetric is an equivalence relation and determine the equivalence classes.

8:38 *** TEMU TEMU -3 -2 7 B 2 1 & 5G. 61% 1 2 -1 Based on the graph above, determine the amplitude, period, midline, and equation of the function. Use f(x) as the output. Amplitude: 2 Period: 2 Midline: 2 ☑ syntax error: this is not an equation. Function: f(x) = −2 cos(πx + 2.5π) +2× Question Help: Worked Example 1 ☑ Message instructor Submit Question ||| <

Let X be the relation defined on the power set of the set integers P(Z) by AXB whenever A U B is a finite set of integers. Prove whether or not X is reflexive, symmetric, antisymmetirc or transitive

Chapter 4 Solutions

Finite Mathematics (11th Edition)

Ch. 4.1 - Convert each inequality into an equation by adding...Ch. 4.1 - Prob. 2E Ch. 4.1 - Convert each inequality into an equation by adding...Ch. 4.1 - Prob. 4E Ch. 4.1 - For Exercises 5-8. (a) determine the number of...Ch. 4.1 - Prob. 6E Ch. 4.1 - For Exercises 5-8, (a) determine the number of...Ch. 4.1 - For Exercises 5-8, (a) determine the number of...Ch. 4.1 - Introduce slack variables as necessary, then...Ch. 4.1 - Introduce slack variables as necessary, then write...

Ch. 4.1 - Introduce slack variables as necessary, then write...Ch. 4.1 - Introduce slack variables as necessary, then write...Ch. 4.1 - Prob. 13E Ch. 4.1 - Introduce slack variables as necessary, then write...Ch. 4.1 - Write the solutions that can be read from each...Ch. 4.1 - Write the solutions that can be read from each...Ch. 4.1 - Prob. 17E Ch. 4.1 - Prob. 18E Ch. 4.1 - Pivot once as indicated in each simplex tableau....Ch. 4.1 - Pivot once as indicated in each simplex tableau....Ch. 4.1 - Prob. 21E Ch. 4.1 - Prob. 22E Ch. 4.1 - Prob. 23E Ch. 4.1 - Prob. 24E Ch. 4.1 - Explain the purpose of a slack variable.Ch. 4.1 - Prob. 26E Ch. 4.1 - Prob. 27E Ch. 4.1 - Set up Exercises 2731 for solution by the simplex...Ch. 4.1 - Set up Exercises 2731 for solution by the simplex...Ch. 4.1 - Set up Exercises 2731 for solution by the simplex...Ch. 4.1 - Set up Exercises 27for solution by the simplex...Ch. 4.2 - In Exercises 1-6, the initial tableau of a linear...Ch. 4.2 - In Exercises 1-6, the initial tableau of a linear...Ch. 4.2 - In Exercises 1-6, the initial tableau of a linear...Ch. 4.2 - In Exercises 1-6, the initial tableau of a linear...Ch. 4.2 - In Exercises 1-6, the initial tableau of a linear...Ch. 4.2 - In Exercises 1- the initial tableau of a linear...Ch. 4.2 - Use the simplex method to solve each linear...Ch. 4.2 - Prob. 8E Ch. 4.2 - Prob. 9E Ch. 4.2 - Use the simplex method to solve each linear...Ch. 4.2 - Prob. 11E Ch. 4.2 - Prob. 12E Ch. 4.2 - Use the simplex method to solve each linear...Ch. 4.2 - Prob. 14E Ch. 4.2 - Use the simplex method to solve each linear...Ch. 4.2 - Prob. 16E Ch. 4.2 - Prob. 17E Ch. 4.2 - Prob. 18E Ch. 4.2 - The simplex algorithm still works if an indicator...Ch. 4.2 - 20. What goes wrong if a quotient other than...Ch. 4.2 - Prob. 21E Ch. 4.2 - Prob. 22E Ch. 4.2 - Prob. 23E Ch. 4.2 - Set up and solve Exercises 23–29 by the...Ch. 4.2 - Set up and solve Exercises 23–29 by the simplex...Ch. 4.2 - Set up and solve Exercises 23–29 by the simplex...Ch. 4.2 - Set up and solve Exercises 2329 by the simplex...Ch. 4.2 - Set up and solve Exercises 2329 by the simplex...Ch. 4.2 - Prob. 29E Ch. 4.2 - Profit A manufacturer makes two products, toy...Ch. 4.2 - Exercises and 32 come from past CPA examinations....Ch. 4.2 - Prob. 32E Ch. 4.2 - Prob. 33E Ch. 4.2 - Prob. 34E Ch. 4.2 - Prob. 35E Ch. 4.2 - Prob. 36E Ch. 4.2 - 37. Resource Management The average weights of...Ch. 4.2 - Prob. 38E Ch. 4.2 - Prob. 39E Ch. 4.2 - Prob. 40E Ch. 4.3 - Find the transpose of each matrix. 1. Ch. 4.3 - Prob. 2E Ch. 4.3 - Prob. 3E Ch. 4.3 - Find the transpose of each matrix. [...Ch. 4.3 - State the dual problem for each linear...Ch. 4.3 - Prob. 6E Ch. 4.3 - State the dual problem for each linear...Ch. 4.3 - Prob. 8E Ch. 4.3 - Prob. 9E Ch. 4.3 - Use the simples method to solve. 10. Find y1 ≥...Ch. 4.3 - Use the simplex method to solve. 11. Find y1 ≥...Ch. 4.3 - Use the simplex method to solve. 12. Minimize w...Ch. 4.3 - Prob. 13E Ch. 4.3 - Use the simplex method to solve. 14. Minimize w =...Ch. 4.3 - Prob. 15E Ch. 4.3 - Prob. 16E Ch. 4.3 - Prob. 17E Ch. 4.3 - Prob. 18E Ch. 4.3 - 19. Production Costs A brewery produces regular...Ch. 4.3 - Supply Costs The chemistry department at a local...Ch. 4.3 - Prob. 21E Ch. 4.3 - Prob. 22E Ch. 4.3 - Prob. 23E Ch. 4.3 - 24. Animal Food An animal food must provide at...Ch. 4.3 - Feed Costs Refer to Example 5 in this section on...Ch. 4.3 - Prob. 26E Ch. 4.3 - Calorie Expenditure Maureen Rachford wants to...Ch. 4.3 - 28. Health Care Greg Conover takes vitamin...Ch. 4.3 - 29. Blending Nutrients A biologist must make a...Ch. 4.4 - Prob. 1E Ch. 4.4 - Prob. 2E Ch. 4.4 - Prob. 3E Ch. 4.4 - Prob. 4E Ch. 4.4 - Convert each problem into a maximization...Ch. 4.4 - Prob. 6E Ch. 4.4 - Prob. 7E Ch. 4.4 - Prob. 8E Ch. 4.4 - Prob. 9E Ch. 4.4 - Prob. 10E Ch. 4.4 - Use the simplex method to solve. Find x1 0, x2 0...Ch. 4.4 - Prob. 12E Ch. 4.4 - Prob. 13E Ch. 4.4 - Prob. 14E Ch. 4.4 - Prob. 15E Ch. 4.4 - Prob. 16E Ch. 4.4 - Prob. 17E Ch. 4.4 - Prob. 18E Ch. 4.4 - Solve using artificial variables. Maximize w =...Ch. 4.4 - Prob. 20E Ch. 4.4 - Prob. 21E Ch. 4.4 - Prob. 22E Ch. 4.4 - Prob. 23E Ch. 4.4 - Prob. 24E Ch. 4.4 - Prob. 25E Ch. 4.4 - Prob. 26E Ch. 4.4 - Prob. 27E Ch. 4.4 - Blending Seed Topgrade Turf lawn seed mixture...Ch. 4.4 - Prob. 29E Ch. 4.4 - Prob. 30E Ch. 4.4 - 31. Blending Chemicals Natural Brand plant food is...Ch. 4.4 - Prob. 32E Ch. 4.4 - 33. Calorie Expenditure Joe Vetere’s exercise...Ch. 4 - Suppose you plan to build a raised flower bed...Ch. 4 - For the list of standard paper roll widths given...Ch. 4 - Prob. 3EA Ch. 4 - For the computer solution of the cutting problem,...Ch. 4 - Prob. 5EA Ch. 4 - Prob. 6EA Ch. 4 - Suppose that of the original 33 cutting patterns,...Ch. 4 - Prob. 8EA Ch. 4 - Compare your two answers from Exercise 8 with each...Ch. 4 - Prob. 10EA Ch. 4 - Determine whether each of the following...Ch. 4 - Prob. 2RE Ch. 4 - Determine whether each of the following...Ch. 4 - Prob. 4RE Ch. 4 - Determine whether each of the following...Ch. 4 - Prob. 6RE Ch. 4 - Prob. 7RE Ch. 4 - Prob. 8RE Ch. 4 - Prob. 9RE Ch. 4 - Determine whether each of the following statements...Ch. 4 - Determine whether each of the following statements...Ch. 4 - Prob. 12RE Ch. 4 - Prob. 13RE Ch. 4 - Prob. 14RE Ch. 4 - Prob. 15RE Ch. 4 - Determine whether each of the following...Ch. 4 - For each problem. (a) add slack variables or...Ch. 4 - For each problem, (a) add slack variables or...Ch. 4 - For each problem, (a) add slack variables or...Ch. 4 - For each problem, (a) add slack variables or...Ch. 4 - Use the simplex method to solve each maximization...Ch. 4 - Use the simplex method to solve each maximization...Ch. 4 - Use the simplex method to solve each maximization...Ch. 4 - Use the simplex method to solve each maximization...Ch. 4 - Convert each problem into a maximization problem...Ch. 4 - Convert each problem into a maximization problem...Ch. 4 - Prob. 27RE Ch. 4 - Prob. 28RE Ch. 4 - Prob. 29RE Ch. 4 - Prob. 30RE Ch. 4 - Prob. 31RE Ch. 4 - Prob. 32RE Ch. 4 - Prob. 33RE Ch. 4 - Prob. 34RE Ch. 4 - Prob. 35RE Ch. 4 - 36. In Chapter 2 we wrote a system of It near...Ch. 4 - Prob. 37RE Ch. 4 - Prob. 38RE Ch. 4 - Prob. 39RE Ch. 4 - Prob. 40RE Ch. 4 - Prob. 41RE Ch. 4 - Prob. 42RE Ch. 4 - Prob. 43RE Ch. 4 - Prob. 44RE Ch. 4 - Prob. 45RE Ch. 4 - Prob. 46RE Ch. 4 - Prob. 47RE

Knowledge Booster

Background pattern image

Math

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, subject and related others by exploring similar questions and additional content below.

Similar questions

SEE MORE QUESTIONS

Recommended textbooks for you

Glencoe Algebra 1, Student Edition, 9780079039897...
Algebra
ISBN:9780079039897
Author:Carter
Publisher:McGraw Hill
College Algebra (MindTap Course List)
Algebra
ISBN:9781305652231
Author:R. David Gustafson, Jeff Hughes
Publisher:Cengage Learning
College Algebra
Algebra
ISBN:9781337282291
Author:Ron Larson
Publisher:Cengage Learning
Big Ideas Math A Bridge To Success Algebra 1: Stu...
Algebra
ISBN:9781680331141
Author:HOUGHTON MIFFLIN HARCOURT
Publisher:Houghton Mifflin Harcourt
Algebra for College Students
Algebra
ISBN:9781285195780
Author:Jerome E. Kaufmann, Karen L. Schwitters
Publisher:Cengage Learning
Algebra & Trigonometry with Analytic Geometry
Algebra
ISBN:9781133382119
Author:Swokowski
Publisher:Cengage

Text book image

Glencoe Algebra 1, Student Edition, 9780079039897...

Algebra

ISBN:9780079039897

Author:Carter

Publisher:McGraw Hill

Text book image

College Algebra (MindTap Course List)

Algebra

ISBN:9781305652231

Author:R. David Gustafson, Jeff Hughes

Publisher:Cengage Learning

Text book image

College Algebra

Algebra

ISBN:9781337282291

Author:Ron Larson

Publisher:Cengage Learning

Text book image

Big Ideas Math A Bridge To Success Algebra 1: Stu...

Algebra

ISBN:9781680331141

Author:HOUGHTON MIFFLIN HARCOURT

Publisher:Houghton Mifflin Harcourt

Text book image

Algebra for College Students

Algebra

ISBN:9781285195780

Author:Jerome E. Kaufmann, Karen L. Schwitters

Publisher:Cengage Learning

Text book image

Algebra & Trigonometry with Analytic Geometry

Algebra

ISBN:9781133382119

Author:Swokowski

Publisher:Cengage

SEE MORE TEXTBOOKS

Use of ALGEBRA in REAL LIFE; Author: Fast and Easy Maths !;https://www.youtube.com/watch?v=9_PbWFpvkDc;License: Standard YouTube License, CC-BY

Compound Interest Formula Explained, Investment, Monthly & Continuously, Word Problems, Algebra; Author: The Organic Chemistry Tutor;https://www.youtube.com/watch?v=P182Abv3fOk;License: Standard YouTube License, CC-BY

Applications of Algebra (Digit, Age, Work, Clock, Mixture and Rate Problems); Author: EngineerProf PH;https://www.youtube.com/watch?v=Y8aJ_wYCS2g;License: Standard YouTube License, CC-BY