...Exercise 3.1 1. Find the minimum and maximum number of items that Jerry can purchase 50≤3x + 5≤80 50-5≤3x+5-5≤80-5 45≤3x≤75 45/3≤3x/3≤75/3 15≤x≤25 The minimum items Jerry can purchase is 15 items The maximum items Jerry can purchase is 25 items 2. Find the maximum horizontal distance that the projectile may cover f(x) = -x2 +2x +2 x =(-2+ 4+8)/-2 x = (2-2 3)/2 <0 not the solution because the distance should be positive x = 1- 3 m is not solution x = (2+2 3)/2>0 is the solution x =1+ 3 m The projectile may cover approximately 2.73m 3. Find the maximum height of the arrow f(x) = -16x2+ 200x+4 f (-200/-32) =f (25/4) = -16(25/4)2+200(25/4) +4 =-625 + 1250 +4 =629 f(25/4) = 629 The maximum height of the arrow is 629 feet 4. Find the number of customers that arrived in the 6th hour f(x) = 2x+1 f(6) = 26+1 = 27 = 128 f(6) = 128 The number of customer that arrived in the 6th hour is 128 customers 5. Find the number of items sold P(x) =3800000 x2- 4000x + 7800000 = 3800000 x2-4000x +7800000-3800000=0 x2-4000x+4000000=0 x= (4000 +- (-4.103)2-4*4.106)/2 x = 2000 The number of items sold is 2000 items 6. Find the value of the machine after 3years f(3) = 20000(1/2)3 = 20000/8 = 2500 f(3) = 2500 After 3 years, the machine will cost $2,500 7. Find the maximum...
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...1) The minimum and maximum number of items that he can purchase is Given: C=3x+5 X= (c-5)/3 Min: X= (50-5)/3 X=15 items minimum Max: x = (80 - 5)/3 x = 25 items maximum 2) Find the maximum horizontal distance that the projectile may cover. Using the quadratic formula Maximum distance of projectile 3) The maximum height of the arrow is: feet Answer is 629 Feet 4) Find the number of customers that arrived in the 6th hour. f(6) = 2^7 = 128 ______________ _______________ _______________ 64 customers arrived in the 6th hr. 5) If the net profit is 3800000, find the number of items sold. P(x) =x^2-4000x + 7800000 3800000=x^2-4000x + 7800000 X^2-4000x+4000000=0 (x-2000) ^2=0 x=2000 P (2000) =3800000 Number of items sold=2000 6) Find its value after 3 years f(x) = 20000(1/2)^x 20,000(.5)^3 20,000*.125=2,500 ans. The value of the Machine after 3 years is $2,500.00 7) Find the maximum height that the object may reach. f(x)=-x2+3x+6 use -b/2a -3/2(-1)=-3/-2=3/2 x=3/2 f(3/2)=-(3/2)2+(3)(3/2)+6 f(3/2)=-(9/4)+(9/2)+6 f(3/2)=-(9/4)+(18/4)+6 f(3/2)=(9/4)+6 f(3/2)=(2 1/4)+6 f(3/2)=8 1/4 meters or 8.25 meters 8) The sum of two numbers is 210. If one number is the square of the other, find the set of numbers. x +y = 210 y = x^2 substitute for y in first equation x^2 +x -210 = 0 factor equation (x+15)*(x-14) = 0 x = -15 or 14 y = 225 or 196 solution1...
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...1. If 5 times the value of a number is increased by 3, the result is 28. What is the number? 5x + 3 = 28 -3 -3 5x = 25 -------- 5 X = 5 The number is 5. 2. If 4 times the reciprocal of a number is more than 5/2 times the reciprocal of that number, find the number. 4 (1/x) = 3 + (5/2) (1/x) 4/x = 3 + 5/2x 2x (4/x) = 2x (3+5/2x) 8 = 6x + 5 -5 - 5 3 = 6x -------- 6 X = ½ The number is ½. 3. John had $30,000 to invest. He invested part of this money in bonds paying 12% annual simple interest and the rest of the money in a savings account giving 4% annual interest. At the end of the year, he received $2,400 as extra income. How much money did John place in each investment? Bonds = x Savings = 30,000 – x .12x + .04 (30,000 – x) = 2,400 .12x + 1,200 - .04x = 2,400 .08x + 1,200 = 2,400 -1,200 -1,200 .08x = 1,200 ----------------- .08 X = $15,000 (amount put into bonds) 30,000 – 15,000 = $15,000 (amount put into savings) John put $15,000 into bonds and $15,000 into savings. 4. The length of a football field 180 feet more than its width. If the perimeter of the field is 1,060 feet, find the length of the field. 2w + 2 (180 +w) = 1,060 2w + 360 + 2w = 1,060 4w + 360 = 1,060 -360 -360 4w – 700 ------------ 4 w = 175 175 +180 = 355 The length of the field is 355 feet. 5. At a school choir concert, 256 students are standing in rows. If the number of students in each...
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...School of Information Technology 1. (25 points) The media access control (MAC) address of a network interface is a unique address. Each network interface relating to its MAC fits the criteria of a function because each device has its own, unique MAC address. Describe an everyday situation in your field that is a function. Be sure to explain exactly why the situation is a function. Using a mac address is a direct link from one connection to another, for example going from a PC to a switch, this is an example of a function because one port on the switch only goes directly to one pc each match up. 2. (20 points) Describe a chart or graph you might find related to a computer that fits the criteria for the graph of a function. Make sure to explain in detail why the graph fits the criteria. The graph of g is the graph of f shifted vertically up by 4 units, the functions meet the criterial of graph in the sense that if a vertical line is drawn it will only cut the graph once. 4. (15 points) For any given graph of a function, explain how to find its domain. If a function f does not model data or verbal conditions, its domain is the largest set of real numbers for which the value of is a real number. Exclude from a function’s domain real numbers that cause division by zero and real numbers that result in a square root of a negative number. For example = x2 – 7x contain no division nor square root. Expression x2-7x represents a real number...
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...MA1210 College Mathematics I Quizzes and Exam QUIZ 1 1. Which of the following is the simplified form of the algebraic expression ( ) a. b. c. d. 2. Which of the following is the simplified form of ( )( )? a. b. c. d. 3. Assume that variables represent nonnegative real numbers. Which of the following is the simplified form of √ √ ? a. √ b. √ c. √ d. √ 4. Evaluate the following algebraic expression for the given values of the variables. ( ) for and a. 148 b. 134 c. 128 d. 142 5. Which of the following is the simplified expression for ( )– ? a. [9] 02/18/2014 MA1210 College Mathematics I Quizzes and Exam b. c. d. – 6. Find the sum of the polynomials ( – ) (– – ) a. – b. – c. – – d. – 7. Find the product of ( ) ( – ). a. b. – c. – d. – – – 8. Factor the following trinomial or state that it is prime. – a. ( )( ) b. ( – )( – ) c. ( – )( ) d. Prime 9. Find all numbers that must be excluded from the domain of the following rational expression: – – a. b. c. – d. – [10] 02/18/2014 MA1210 College Mathematics I Quizzes and Exam 10. Find the simplified form of the following rational expression: a. b. c. 1 d. [11] 02/18/2014 MA1210 College Mathematics I Quizzes and Exam QUIZ 1: ANSWER SHEET DATE: STUDENT NAME: COURSE NUMBER: ...
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...ITT Technical Institute IS3340 Windows Security Onsite Course SYLLABUS Credit hours: 4.5 Contact/Instructional hours: 60 (30 Theory Hours, 30 Lab Hours) Prerequisite(s) and/or Corequisite(s): Prerequisite: NT2580 Introduction to Information Security or equivalent Course Description: This course examines security implementations for a variety of Windows platforms and applications. Areas of study include analysis of the security architecture of Windows systems. Students will identify and examine security risks and apply tools and methods to address security issues in the Windows environment. Windows Security Syllabus Where Does This Course Belong? This course is required for the Bachelor of Science in Information Systems Security program. This program covers the following core areas: Foundational Courses Technical Courses BSISS Project The following diagram demonstrates how this course fits in the program: IS4799 NT2799 IS4670 ISC Capstone Project Capstone ProjectCybercrime Forensics NSA NT2580 NT2670 Introduction to Information Security IS4680 IS4560 NT2580 NT2670 Email and Web Services Hacking and Introduction to Security Auditing for Compliance Countermeasures Information Security Email and Web Services NT1230 NT1330 Client-Server Client-Server Networking I Networking II IS3230 IS3350 NT1230 NT1330 Issues Client-Server Client-Server SecurityContext in Legal Access Security Networking I Networking II NT1110...
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...or Corerequisite: NT1210 Introduction to Networking or equivalent Course Description: This course introduces operating principles for the client-server based networking systems. Students will examine processes and procedures involving the installation, configuration, maintanence, troublshooting and routine adminstrative tasks of popular desktop operating system(s) for standalone and network client computers, and related aspects of typical network server functions. Client-Server Networking I Syllabus Where Does This Course Belong? 1st QTR GS1140 NT1110 GS1145 Problem Solving Theory Computer Structure and Logic Strategies for the Technical Professional 2nd QTR NT1210 Introduction to Networking NT1230 Client-Server Networking I MA1210 College Mathematics I 3rd QTR NT1310 NT1330 MA1310 4th QTR PT1420 NT1430 EN1320 5th QTR PT2520 NT2580 EN1420 6th QTR NT2640 NT2670 CO2520 7th QTR NT2799 SP2750 Physical Networking Client-Server Networking II College Mathematics II Introduction to Programming Linux Networking Composition I Database Concepts Introduction to Information Security Composition II IP Networking Email and Web Services Communications Network Systems Administration Capstone Project Group Theory The follow diagram indicates how this course relates to other courses in the NSA program: 1 Date: 8/31/2012 Client-Server Networking I Syllabus NT2799 NSA Capstone Project NT2580 Introduction to Information Security NT2670 Email and Web Services ...
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...MA1210 Module 3 exercise 1) 3x+5=50 3x+5-5=50-5 3x=45 3x/3=45/3 X=15 3x+5=80 3x+5-5=80-5 3x/3=75/3 X=25 2) F(x)=-x^2+2x+2 X=-b+/-sqrt(b^2-4ac)/2a X=-2+/-sqrt(2^2-4*-1*2)/2*-1 X=-2+/-sqrt(4+8)/-2 3-(-4) 3+4 2^2-4*-1*2 4-(-8) 4+8 X=-2+/--sqrt(4+8)/-2 X=-2+/-sqrt(12)/-2 Sqrt12 Sqrt(4*3) 2sqrt(3) X=-2+/-SQRT(12)/-2 X=-2+2sqrt(3)/-2 2(-1+sqrt 3)/-2 -1+/- sqrt (3)/-1 3) f(x)=16^2+200x+4 X=b/2a X=200/2*-16 X=-200/-32 X=6.25 F(x)=-16x^2+200x+4 F(6.25)=-16*(6.25)^2+200*6.25+4 =16*39.0625+1250+4 =-625+1250+4 =-625+1254 = 629 feet 4) F(x)=2^x+1 6=2^x+1 6=2^7 6=128 5) P(x)=x^2-4000x+7,800,000 X^2-4000x+7,800,000-3,800,000=3,800,000-3,800,000 X^2-4000x+4,000,000=0 X^2-2*2000*x+(2000)^2=0 (x-2000)^2=0 x-2000=0 x-2000+2000=0+2000 x=2000 6) F(x)=20,000(1/2)^x 3=20,000(1/2)^3 3=20,000*1/8 3=2500 7) F(x)=-x^2+3x+6 X=-b/2a X=-3/2*-1 X=-3/-2 X=1.5 1.5=-1.5^2+3*1.5+6 =-2.25+4.5+6 =-2.25+10.5 =8.25 8) X+y=210 X=y^2 Y^2+y=210 Y^2+y-210=210-210 Y^2+y-210=0 (y+15)(y-14)=0 Y+15=0 Y+15-15=0-15 Y=-15 y-14+14=0+14 y=14 x+y=210 x-15=210 x+15+15=210+15 x=225 (225,15) X+y=210 X+14=210 X+14-14=210-14 X=196 (194,14) (225,15),(196,14) 9) 12=x*y X=5+2*y X=5+2y 12=(5+2y)*y 12=5*y+2y*5 12=5*y+2y*y 12=5y+2y^2 2y^2+5y=12 2y^2+5y-12=12-2 2y^2+5-12=0 2*12=24 24=2*2*2*3 (8.3) (2y-3)(y+4) 2y*y+3y*4-3*y-3*4 2y^2+8y-3y-12 2y^2+5y-12=0 (2y-3)(y+4)=0 2y-3=0 2y-3+3=0+3 2y=3 2y/2=3/2 Y=3/2 Y+4=0 ...
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...Communications Technology. The following diagrams indicate how this course relates to other courses in respective programs: Network Systems Administration NT2799 NSA Capstone Project NT2580 Introduction to Information Security NT2670 Email and Web Services NT2640 IP Networking PT2520 Database Concepts NT1330 Client-Server Networking II NT1230 Client-Server Networking I NT1430 Linux Networking PT1420 Introduction to Programming NT1110 Computer Structure and Logic NT1201 Introduction to Networking NT1310 Physical Networking CO2520 Communications SP2750 Group Theories EN1420 Composition II EN1320 Composition I GS1140 Problem Solving Theory GS1145 Strategies for the Technical Professional MA1210 College Mathematics I MA1310 College Mathematics II Networking Technology Courses Programming Technology Courses General Education/ General Studies 1 Date: 7/18/2011 IP Networking Syllabus Mobile Communications Technology MC2799 MCT Capstone Project NT2640 IP Networking MC2560 Mobile Wireless Communication I MC2660 Mobile Wireless Communication II MC2665 Mobile...
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...Electrical Engineering Technology. The following diagrams demonstrate how this course fits in each program. Associate Program in Network Systems Administration NT2799 NSA Capstone Project NT2580 Introduction to Information Security NT2670 Email and Web Services NT2640 IP Networking PT2520 Database Concepts NT1330 Client-Server Networking II NT1230 Client-Server Networking I NT1430 Linux Networking PT1420 Introduction to Programming NT1110 Computer Structure and Logic NT1210 Introduction to Networking NT1310 Physical Networking CO2520 Communications SP2750 Group Theories EN1420 Composition II EN1320 Composition I GS1140 Problem Solving Theory GS1145 Strategies for the Technical Professional MA1210 College Mathematics I MA1310 College Mathematics II Networking Technology Courses Programming...
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...Electrical Engineering Technology. The following diagrams demonstrate how this course fits in each program. Associate Program in Network Systems Administration NT2799 NSA Capstone Project NT2580 Introduction to Information Security NT2670 Email and Web Services NT2640 IP Networking PT2520 Database Concepts NT1330 Client-Server Networking II NT1230 Client-Server Networking I NT1430 Linux Networking PT1420 Introduction to Programming NT1110 Computer Structure and Logic NT1210 Introduction to Networking NT1310 Physical Networking CO2520 Communications SP2750 Group Theories EN1420 Composition II EN1320 Composition I GS1140 Problem Solving Theory GS1145 Strategies for the Technical Professional MA1210 College Mathematics I MA1310 College Mathematics II Networking Technology Courses Programming...
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...ITT Technical Institute NT1310 Physical Networking Student Course Package Bring this document with you each week Students are required to complete each assignment and lab in this course package on time whether or not they are in class. Late penalties will be assessed for any assignments or labs handed in past the due date. The student is responsible for replacement of the package if lost. Table of Contents Syllabus 2 Student Professional Experience 19 Graded Assignments and Exercises 23 Labs 47 Documenting your Student Professional Experience 57 ITT Technical Institute NT1310 Physical Networking Onsite Course SYLLABUS Credit hours: 4.5 Contact/Instructional hours: 56 (34 Theory Hours, 22 Lab Hours) Prerequisite(s) and/or Corequisite(s): Prerequisites: NT1210 Introduction to Networking or equivalent Course Description: This course examines industry standards and practices involving the physical components of networking technologies (such as wiring standards and practices, various media and interconnection components), networking devices and their specifications and functions. Students will practice designing physical network solutions based on appropriate capacity planning and implementing various installation, testing and troubleshooting techniques for a computer network. Where Does This Course Belong? | | | NT2799 | | | | | | | | NSA Capstone | | | | | | | Project | | | | | NT2580...
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