
Pearson eText for Business Math -- Instant Access (Pearson+)
11th Edition
ISBN: 9780137518883
Author: Cheryl Cleaves, Margie Hobbs
Publisher: PEARSON+
expand_more
expand_more
format_list_bulleted
Concept explainers
Question
Chapter 1.2, Problem 42SE
To determine
Whether the bid is low bid for the job and also tells if it will likely to get the business if the area of a land is square piece of land is 210 feet on each of the 4 equal side and fencing can be buy in 50-foot rolls for $49 per roll.
Expert Solution & Answer

Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
B:Show that the function 4H(x,y)= (x² + y2)2-2((x² + y²) is a first
integral of ODES:
x=y + y(x² + y²)
y=x+x (x² + y²)
and sketch the stability of critical points and draw the phase portrait of
system.
A: Show that the ODES has no limit cycle in a region D and find this region:
x=y-2x³
y=x+y-2y3
C
optımızatıon theory
Chapter 1 Solutions
Pearson eText for Business Math -- Instant Access (Pearson+)
Ch. 1.1 - Prob. 1-1SCCh. 1.1 - Prob. 1-2SCCh. 1.1 - Prob. 1-3SCCh. 1.1 - Prob. 1-4SCCh. 1.1 - Prob. 2-1SCCh. 1.1 - Prob. 2-2SCCh. 1.1 - Prob. 2-3SCCh. 1.1 - Prob. 2-4SCCh. 1.1 - Prob. 3-1SCCh. 1.1 - Prob. 3-2SC
Ch. 1.1 - Prob. 3-3SCCh. 1.1 - Prob. 3-4SCCh. 1.1 - Prob. 3-5SCCh. 1.1 - Prob. 3-6SCCh. 1.1 - Prob. 4-1SCCh. 1.1 - Prob. 4-2SCCh. 1.1 - Prob. 4-3SCCh. 1.1 - Prob. 4-4SCCh. 1.1 - Prob. 1SECh. 1.1 - Prob. 2SECh. 1.1 - Prob. 3SECh. 1.1 - Prob. 4SECh. 1.1 - Prob. 5SECh. 1.1 - Prob. 6SECh. 1.1 - Prob. 7SECh. 1.1 - Prob. 8SECh. 1.1 - Prob. 9SECh. 1.1 - Prob. 10SECh. 1.1 - Prob. 11SECh. 1.1 - Prob. 12SECh. 1.1 - Prob. 13SECh. 1.1 - Prob. 14SECh. 1.1 - Prob. 15SECh. 1.1 - Prob. 16SECh. 1.1 - Prob. 17SECh. 1.1 - Prob. 18SECh. 1.1 - Prob. 19SECh. 1.1 - Prob. 20SECh. 1.1 - Prob. 21SECh. 1.1 - Prob. 22SECh. 1.1 - Prob. 23SECh. 1.1 - Prob. 24SECh. 1.1 - Prob. 25SECh. 1.1 - Prob. 26SECh. 1.2 - Prob. 1-1SCCh. 1.2 - Prob. 1-2SCCh. 1.2 - Prob. 1-3SCCh. 1.2 - Prob. 1-4SCCh. 1.2 - Prob. 1-5SCCh. 1.2 - Prob. 1-6SCCh. 1.2 - Prob. 1-7SCCh. 1.2 - Prob. 1-8SCCh. 1.2 - Prob. 1-9SCCh. 1.2 - Prob. 1-10SCCh. 1.2 - Prob. 1-11SCCh. 1.2 - Prob. 1-12SCCh. 1.2 - Prob. 2-1SCCh. 1.2 - Prob. 2-2SCCh. 1.2 - Prob. 2-3SCCh. 1.2 - Prob. 2-4SCCh. 1.2 - Prob. 2-5SCCh. 1.2 - Prob. 2-6SCCh. 1.2 - Prob. 3-1SCCh. 1.2 - Prob. 3-2SCCh. 1.2 - Prob. 3-3SCCh. 1.2 - Prob. 3-4SCCh. 1.2 - Prob. 3-5SCCh. 1.2 - Prob. 3-6SCCh. 1.2 - Prob. 3-7SCCh. 1.2 - Prob. 3-8SCCh. 1.2 - Prob. 3-9SCCh. 1.2 - Prob. 3-10SCCh. 1.2 - Prob. 4-1SCCh. 1.2 - Prob. 4-2SCCh. 1.2 - Prob. 4-3SCCh. 1.2 - Prob. 4-4SCCh. 1.2 - Prob. 4-5SCCh. 1.2 - Prob. 4-6SCCh. 1.2 - Prob. 4-7SCCh. 1.2 - Prob. 4-8SCCh. 1.2 - Prob. 4-9SCCh. 1.2 - Prob. 4-10SCCh. 1.2 - Prob. 5-1SCCh. 1.2 - Prob. 5-2SCCh. 1.2 - Prob. 5-3SCCh. 1.2 - Prob. 5-4SCCh. 1.2 - Prob. 1SECh. 1.2 - Prob. 2SECh. 1.2 - Prob. 3SECh. 1.2 - Prob. 4SECh. 1.2 - Prob. 5SECh. 1.2 - Prob. 6SECh. 1.2 - Prob. 7SECh. 1.2 - Prob. 8SECh. 1.2 - Prob. 9SECh. 1.2 - Prob. 10SECh. 1.2 - Prob. 11SECh. 1.2 - Prob. 12SECh. 1.2 - Prob. 13SECh. 1.2 - Prob. 14SECh. 1.2 - Prob. 15SECh. 1.2 - Prob. 16SECh. 1.2 - Prob. 17SECh. 1.2 - Prob. 18SECh. 1.2 - Prob. 19SECh. 1.2 - Prob. 20SECh. 1.2 - Prob. 21SECh. 1.2 - Prob. 22SECh. 1.2 - Prob. 23SECh. 1.2 - Prob. 24SECh. 1.2 - Prob. 25SECh. 1.2 - Prob. 26SECh. 1.2 - Prob. 27SECh. 1.2 - Prob. 28SECh. 1.2 - Prob. 29SECh. 1.2 - Prob. 30SECh. 1.2 - Prob. 31SECh. 1.2 - Prob. 32SECh. 1.2 - Prob. 33SECh. 1.2 - Prob. 34SECh. 1.2 - Prob. 35SECh. 1.2 - Prob. 36SECh. 1.2 - Prob. 37SECh. 1.2 - Prob. 38SECh. 1.2 - Prob. 39SECh. 1.2 - Prob. 40SECh. 1.2 - Prob. 41SECh. 1.2 - Prob. 42SECh. 1.2 - Prob. 43SECh. 1.2 - Prob. 44SECh. 1.2 - Prob. 45SECh. 1.2 - Prob. 46SECh. 1.2 - Prob. 47SECh. 1.2 - Prob. 48SECh. 1.2 - Prob. 49SECh. 1.2 - Prob. 50SECh. 1.2 - Prob. 51SECh. 1.2 - Prob. 52SECh. 1.2 - Prob. 53SECh. 1.2 - Prob. 54SECh. 1.2 - Prob. 55SECh. 1.2 - Prob. 56SECh. 1 - Prob. 1ESCh. 1 - Prob. 2ESCh. 1 - Prob. 3ESCh. 1 - Prob. 4ESCh. 1 - Prob. 5ESCh. 1 - Prob. 6ESCh. 1 - Prob. 7ESCh. 1 - Prob. 8ESCh. 1 - Prob. 9ESCh. 1 - Prob. 10ESCh. 1 - Prob. 11ESCh. 1 - Prob. 12ESCh. 1 - Prob. 13ESCh. 1 - Prob. 14ESCh. 1 - Prob. 15ESCh. 1 - Prob. 16ESCh. 1 - Prob. 17ESCh. 1 - Prob. 18ESCh. 1 - Prob. 19ESCh. 1 - Prob. 20ESCh. 1 - Prob. 21ESCh. 1 - Prob. 22ESCh. 1 - Prob. 23ESCh. 1 - Prob. 24ESCh. 1 - Prob. 25ESCh. 1 - Prob. 26ESCh. 1 - Prob. 27ESCh. 1 - Prob. 28ESCh. 1 - Prob. 29ESCh. 1 - Prob. 30ESCh. 1 - Prob. 31ESCh. 1 - Prob. 32ESCh. 1 - Prob. 33ESCh. 1 - Prob. 34ESCh. 1 - Prob. 35ESCh. 1 - Prob. 36ESCh. 1 - Prob. 37ESCh. 1 - Prob. 38ESCh. 1 - Prob. 39ESCh. 1 - Prob. 40ESCh. 1 - Prob. 41ESCh. 1 - Prob. 42ESCh. 1 - Prob. 43ESCh. 1 - Prob. 44ESCh. 1 - Prob. 45ESCh. 1 - Prob. 46ESCh. 1 - Prob. 47ESCh. 1 - Prob. 48ESCh. 1 - Prob. 49ESCh. 1 - Prob. 50ESCh. 1 - Prob. 51ESCh. 1 - Prob. 52ESCh. 1 - Prob. 53ESCh. 1 - Prob. 54ESCh. 1 - Prob. 55ESCh. 1 - Prob. 56ESCh. 1 - Prob. 57ESCh. 1 - Prob. 58ESCh. 1 - Prob. 59ESCh. 1 - Prob. 60ESCh. 1 - Prob. 61ESCh. 1 - Prob. 62ESCh. 1 - Prob. 63ESCh. 1 - Prob. 64ESCh. 1 - Prob. 65ESCh. 1 - Prob. 66ESCh. 1 - Prob. 67ESCh. 1 - Prob. 68ESCh. 1 - Prob. 69ESCh. 1 - Prob. 70ESCh. 1 - Prob. 71ESCh. 1 - Prob. 72ESCh. 1 - Prob. 73ESCh. 1 - Prob. 74ESCh. 1 - Prob. 75ESCh. 1 - Prob. 76ESCh. 1 - Prob. 77ESCh. 1 - Prob. 78ESCh. 1 - Prob. 79ESCh. 1 - Prob. 80ESCh. 1 - Prob. 81ESCh. 1 - Prob. 82ESCh. 1 - Prob. 83ESCh. 1 - Prob. 84ESCh. 1 - Prob. 85ESCh. 1 - Prob. 86ESCh. 1 - Prob. 87ESCh. 1 - Prob. 88ESCh. 1 - Prob. 89ESCh. 1 - Prob. 1PTCh. 1 - Prob. 2PTCh. 1 - Prob. 3PTCh. 1 - Prob. 4PTCh. 1 - Prob. 5PTCh. 1 - Prob. 6PTCh. 1 - Prob. 7PTCh. 1 - Prob. 8PTCh. 1 - Prob. 9PTCh. 1 - Prob. 10PTCh. 1 - Prob. 11PTCh. 1 - Prob. 12PTCh. 1 - Prob. 13PTCh. 1 - Prob. 14PTCh. 1 - Prob. 15PTCh. 1 - Prob. 16PTCh. 1 - Prob. 17PTCh. 1 - Prob. 18PTCh. 1 - Prob. 19PTCh. 1 - Prob. 20PTCh. 1 - Prob. 21PTCh. 1 - Prob. 22PTCh. 1 - Prob. 23PTCh. 1 - Prob. 24PTCh. 1 - Prob. 25PTCh. 1 - Prob. 26PTCh. 1 - Prob. 27PTCh. 1 - Prob. 28PTCh. 1 - Prob. 29PTCh. 1 - Prob. 30PTCh. 1 - Prob. 31PTCh. 1 - Prob. 32PTCh. 1 - Prob. 1CTCh. 1 - Prob. 2CTCh. 1 - Prob. 3CTCh. 1 - Prob. 4CTCh. 1 - Prob. 5CTCh. 1 - Prob. 6CTCh. 1 - Prob. 7CTCh. 1 - Prob. 8CTCh. 1 - Prob. 9CTCh. 1 - Prob. 10CTCh. 1 - Prob. 11CTCh. 1 - Prob. 12CTCh. 1 - Prob. 1CPCh. 1 - Prob. 2CPCh. 1 - Prob. 1CS1Ch. 1 - Prob. 2CS1Ch. 1 - Prob. 3CS1Ch. 1 - Prob. 1CS2Ch. 1 - Prob. 2CS2Ch. 1 - Prob. 3CS2Ch. 1 - Prob. 1CS3Ch. 1 - Prob. 2CS3Ch. 1 - Prob. 3CS3Ch. 1 - Prob. 4CS3Ch. 1 - Prob. 5CS3
Knowledge Booster
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
- Solve : X + sin x = 0. By the false positioning numerical methodarrow_forwardSolve: X + sin X = 0 by the false positionining numerical methodarrow_forwardQ3)A: Given H(x,y)= x²-x4 + y² as a first integral of an ODEs, find this ODES corresponding to H(x,y) and show the phase portrait by using Hartman theorem and by drawing graph of H(x,y)=c. Discuss the stability of critical points of the corresponding ODEs.arrow_forward
- On from the equation: 2 u = C₁ + C₂ Y + Czy + Cu y³ Find C₁, C₂, C3 and Cy Using these following Cases : (a) 4=0 at y=0 (b) U = U∞ at y = 8 du (c) at Y = S ду --y. ди = 0 at y = 0 бугarrow_forwardQI Find the first integral + (x°) ³ + x =0arrow_forwardQ1: solve the system y 2 In √√x² + y2 X y = −y + In √√x² + y2 and solve the linear part.arrow_forward
- Using the method of sections need help solving this please explain im stuckarrow_forwardFind all the solutions of the congruence 7x² + 15x = 4 (mod 111).arrow_forward) The set {1,2,..., 22} is to be split into two disjoint non-empty sets S and T in such a way that: (i) the product (mod 23) of any two elements of S lies in S; (ii) the product (mod 23) of any two elements of T lies in S; (iii) the product (mod 23) of any element of S and any element of T lies in T. Prove that the only solution is S = {1, 2, 3, 4, 6, 8, 9, 12, 13, 16, 18}, T= {5, 7, 10, 11, 14, 15, 17, 19, 20, 21, 22}.arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
- Glencoe Algebra 1, Student Edition, 9780079039897...AlgebraISBN:9780079039897Author:CarterPublisher:McGraw HillAlgebra: Structure And Method, Book 1AlgebraISBN:9780395977224Author:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. ColePublisher:McDougal LittellHolt Mcdougal Larson Pre-algebra: Student Edition...AlgebraISBN:9780547587776Author:HOLT MCDOUGALPublisher:HOLT MCDOUGAL
- Big Ideas Math A Bridge To Success Algebra 1: Stu...AlgebraISBN:9781680331141Author:HOUGHTON MIFFLIN HARCOURTPublisher:Houghton Mifflin HarcourtElementary AlgebraAlgebraISBN:9780998625713Author:Lynn Marecek, MaryAnne Anthony-SmithPublisher:OpenStax - Rice University

Glencoe Algebra 1, Student Edition, 9780079039897...
Algebra
ISBN:9780079039897
Author:Carter
Publisher:McGraw Hill

Algebra: Structure And Method, Book 1
Algebra
ISBN:9780395977224
Author:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Publisher:McDougal Littell

Holt Mcdougal Larson Pre-algebra: Student Edition...
Algebra
ISBN:9780547587776
Author:HOLT MCDOUGAL
Publisher:HOLT MCDOUGAL

Big Ideas Math A Bridge To Success Algebra 1: Stu...
Algebra
ISBN:9781680331141
Author:HOUGHTON MIFFLIN HARCOURT
Publisher:Houghton Mifflin Harcourt


Elementary Algebra
Algebra
ISBN:9780998625713
Author:Lynn Marecek, MaryAnne Anthony-Smith
Publisher:OpenStax - Rice University
Find number of persons in a part with 66 handshakes Combinations; Author: Anil Kumar;https://www.youtube.com/watch?v=33TgLi-wp3E;License: Standard YouTube License, CC-BY
Discrete Math 6.3.1 Permutations and Combinations; Author: Kimberly Brehm;https://www.youtube.com/watch?v=J1m9sB5XZQc;License: Standard YouTube License, CC-BY
How to use permutations and combinations; Author: Mario's Math Tutoring;https://www.youtube.com/watch?v=NEGxh_D7yKU;License: Standard YouTube License, CC-BY
Permutations and Combinations | Counting | Don't Memorise; Author: Don't Memorise;https://www.youtube.com/watch?v=0NAASclUm4k;License: Standard Youtube License
Permutations and Combinations Tutorial; Author: The Organic Chemistry Tutor;https://www.youtube.com/watch?v=XJnIdRXUi7A;License: Standard YouTube License, CC-BY