Technical Mathematics

Transcription

1 Technical Mathematics Author: Todd Simeone (Technology Faculty, Heartland Community College) Volume I Updated November 18, 2016

2 Table of Contents Chapter 1 Prerequisite material Arithmetic with whole numbers Fractions Mixed numbers and Decimals Order of Operations Solving Linear Equations Ratio and Proportion Scientific Notation Chapter 1 Homework problems Chapter 2 Measurement Systems Standard Systems Areas and Volumes Temperature Converting between systems Chapter 2 Homework problems Chapter 3 Percentages, precision, error, and tolerance intervals Percentages: basic conversions Percentages: base, part, and rate Precision and error Tolerance intervals Chapter 3 Homework problems Chapter 4 Number systems Decimal number system Electronic machines Base-2 (Binary) Base 16 (Hexadecimal) Chapter 4 Homework problems Chapter 5 Formulas and data substitution Substituting data into formulas

3 5.2 Simple algebraic manipulation of formulas Chapter 5 Homework problems Chapter 6 Basic Geometry Angles Triangles The Pythagorean Theorem Area Quadrilaterals Circles Circle terminology and basic formulas Arcs and associate angles Chapter 6 Homework problems Chapter 7 Basic Trigonometry The Trigonometric ratios Finding trigonometric values on a calculator Using the calculator to determine trigonometric values of angles Using the calculator to determine angles given their trigonometric values Using trigonometry to find unknown components of a triangle Chapter 7 Homework problems Chapter 8 Interpreting Charts and Graphs Charts and Graphs Bar charts Pie charts Coordinate system graphs Chapter 8 Homework problems Chapter 9 Basic Statistics Data presentation Measures of central tendency Measures of dispersion Normally distributed data sets Controlled and capable processes Chapter 9 Homework problems

4 Chapter 10 Basic Logic Logical expressions The And logical operator The Or logical operator The Not logical operator Logic truth tables Schematics Chapter 10 Homework problems Supplemental Chapter Systems of Linear Equations S.1 Solving systems of 2 linear equations using a graphing tool S.2 Solving systems of 2 equations using the Addition-Subtraction method S.3 Solving systems of 2 equations using the Substitution method S.4 Applications associated with systems of equations S.5 Systems of 3 linear equations Chapter S Homework problems Answers to Homework Problems Chapter 1 Answers Chapter 2 Answers Chapter 3 Answers Chapter 4 Answers Chapter 5 Answers Chapter 6 Answers Chapter 7 Answers Chapter 8 Answers Chapter 9 Answers Chapter 10 Answers Supplemental Chapter Answers

5 Chapter 1 Prerequisite material It is assumed that readers of this text have been exposed to the majority of the Chapter 1 content (if not all of it). As such much of the content is provided with brevity. 1.1 Arithmetic with whole numbers It is assumed that the reader is comfortable with the content of section 1.1 more than any other section in this chapter. A small sampling of examples is provided with no formal discussion. Adding whole numbers: Example ( 14) ( 3) ( 14) 17 Subtracting whole numbers: Example ( 14) ( 3) ( 14) 11 Multiplying whole numbers: Example ( 14) ( 3) ( 14) 42 5

6 Dividing whole numbers: Example ( 3) 5 ( 15) 3 5 ( 15) ( 3) 5 Exponentiation: Example Example Example ,144 6

7 1.2 Fractions Adding and subtracting fractions When adding fractions, the denominator must be the same. This is referred to as a common denominator. As in the first section, only a limited discussion is provided for this review material. Example Since the two fractions had 6 as the common denominator, the two numerators are added and placed over the common denominator A similar process is used to subtract fractions: Example When adding fractions that do not have the same denominator, a common denominator must be found. Example can be used as the common denominator since 7 and 3 are both 7 3 divisors of The fractions in the original problem are replaced with equivalent fractions over that have a common denominator The answer to the original problem is

8 A similar process is used when subtracting fractions with different denominators. Example The answer to the original problem is Mixed numbers and Decimals Mixed numbers are numbers that contain both an integer and a fraction. They always have an equivalent fractional representation knows as an improper fraction. Example 1 Write the improper fraction 23 7 as a mixed number. Solution: When 23 is divided by 7, the result is 3 with a remainder of two. 3 becomes the integer in the mixed number, 2 is the numerator of the fraction, and 7 is the denominator as follows: Example 2 Write the mixed number as an improper fraction. Solution: To determine the numerator, the following calculation is used: (413) As a result, 63 becomes the numerator and the denominator is 13 as follows:

9 In addition to improper fractions and mixed numbers, decimal numbers are another way to represent these values. Example 3 Write the fraction 3 8 as a decimal. Solution: Using a calculator, , and so Example 4 Write the mixed number as a decimal. 7 Solution: Using a calculator, , and so Example 5 Write the decimal 0.2 as a fraction. Solution: Since the 2 is in the tenths place, Example 6 Write the decimal 3.21 as a mixed number. Solution: Since the 1 is in the hundredths place,

10 1.4 Order of Operations When multiple operations are to be performed, there is an accepted order in which the operations must take place. Many students remember the acronym PEMDAS to help keep track of this order. PEMDAS stands for the following: 1. P = Parenthesis Any operations within parentheses are done first. Parentheses are used to override the normal order of operations. 2. E = Exponents 3. MD = Multiplication and division In step 3, all multiplication and division operations are done. If there is more than one, they are done in the order they appear from left to right. 4. AS = Addition and subtraction In step 4, all addition and subtraction operations are done. If there are more than one, they are done in the order they appear from left to right. Example 1 Calculate: 3 33 Solution: is calculated first since multiplication is done before addition is calculated since it is the lone remaining calculation to perform 12 The resulting answer is 12 10

11 Example 2 2 Calculate: (3 4) Solution: (3 4) ( 3 4) is calculated first since it is within parentheses is calculated next since it is an exponent is calculated next since it is the left-most multiplication is calculated next since it is multiplication is calculated next since all remaining operations have the same precedence and it is the left-most operation remaining is calculated next since all remaining operations have the same precedence and it is the left-most operation remaining is the lone remaining calculation to perform 92 The resulting answer is 92 11

12 1.5 Solving Linear Equations Linear equations are equations that have only one variable that has an exponent of 1. For example, the following is a linear equation: x 5 14 The goal when solving a linear equation is to determine all of the numbers that can be substituted in place of the variable which result in a true statement. 9 is a solution to the above equation because when 9 is substituted in place of x, a true statement results: Example 1a x Following a similar line of thinking, 10 is NOT a solution because when 10 is substituted in place of x, a false statement results: Example 1b x Additional values can be substituted in for x, but only 9 will result in a true statement (and hence only 9 is a solution to this equation). As equations become more complex, there are a few rules that can be applied to help determine the solutions: 1. Any value can be added to or subtracted from both sides of an equation. 2. Both sides of the equation can be multiplied or divided by any non-zero value. 12

13 Example 2 Find the solution of the equation: 3x 17 2 Solution: 3x x is added to both sides of the equation 3x 15 3x x 5 Both sides of the equation are divided by 3 x has been isolated, and the solution is Ratio and Proportion Ratios are ways to compare one measurement to another and are represented as fractions. For example, if the wall in a room is 30m, and the opposite wall is 100m, the values can be represented as: which can be simplified to ratios are in proportion if they are equal to each other. For example, because and are in proportion Sometimes it is necessary to determine a specific value which will put two ratios in proportion. Crossmultiplication can be used to accomplish this task as follows: 13

14 Example 1 Find the value for x which solves the following proportion: Solution: x x x Cross-multiplication is performed on the proportion 31x x Both sides of the equation are divided by 31 x 35 x has been isolated, and the solution is Scientific Notation Scientific notation provides another way to represent numbers. This notation is often used in a technical setting to provide a short-hand notation for numbers that might otherwise be very long. Scientific notation depends on powers of 10 to represent numbers. If you consider the following calculation: the calculated value is: 36,000,000,000,000. Since these are just two different representations of the same value, either can be used. However, since the first one is shorter and more concise it is often used in technical settings, and is in fact known as the scientific notation representation. The second representation is in what is known as decimal notation. It can be seen that the second representation was generated from the first number by simply moving the decimal point 13 places to the right. Scientific notation representations always work this way. The only rules to remember when converting from scientific notation are: 1. If the exponent is positive, move the decimal point to the right 2. If the exponent is negative, move the decimal point to the left 14

15 The following table demonstrates this process: Scientific notation Rule to be applied Converted to decimal notation Positive exponent move decimal to right 47,000, Positive exponent move decimal to right 12,600,000, Negative exponent move decimal to left Negative exponent move decimal to left To convert from decimal notation to scientific notation, the process is reversed. Furthermore, the decimal point is always moved so that the remaining number only has one digit to the left of the decimal point. The rules to convert from decimal notation to scientific notation are as follows: 1. Move the decimal point so the number has exactly one digit to the left of the decimal 2. The exponent is the number of places the decimal point was moved a. If the decimal was moved to the left, the exponent is positive b. If the decimal was moved to the right, the exponent is negative The following table demonstrates this process: Decimal notation Rule to be applied Converted to scientific notation 231,000,000 Move decimal to left positive exponent ,127,000,000,000,000 Move decimal to left positive exponent Move decimal to right negative exponent Move decimal to right negative exponent When performing calculations with numbers in scientific notation, the most straightforward way to handle it is to perform the calculations on the numbers separately from the calculations on the powers of 10 as in the following example: 15

16 Example 1 Calculate the following: 8 5 (3.210 ) ( ) Solution: 8 5 (3.210 ) ( ) 8 5 ( ) (10 10 ) Example 2 Calculate the following: 11 3 ( ) ( ) Solution: 11 3 ( ) ( ) As demonstrated by the next example, care must be taken to ensure the answer is in scientific notation. Example 2 Calculate the following: 5 3 (9.210 ) ( ) Solution: 5 3 (9.210 ) (8.510 ) 5 3 ( ) (10 10 )

17 Chapter 1 Homework problems Section 1.2 Homework Calculate the following

18 Section 1.3 Homework Fill in the following chart. Mixed number Improper Fraction Decimal Section 1.4 Homework Calculate the following (7 2) (7 2) (11 8) (8 3 ) (3 7)

19 Section 1.5 Homework In each of the equations, solve for the variable. 1. x x x4 3x x x x Section 1.6 Homework In each of the following, find the value of x which solves the proportion x x x x x

20 Section 1.7 Homework Fill in the following chart: Scientific Notation Decimal Notation 509,000,000, Calculate the following and express the answer in scientific notation (2 10 ) (4 10 ) ( ) ( ) (310 ) (3 10 ) (5.110 ) ( ) ( ) ( ) ( ) 4 6 (9.110 ) (410 ) 20

21 Chapter 2 Measurement Systems 2.1 Standard Systems While the U.S. typically uses the Imperial System for measurement, the rest of the world uses the Metric System. For this reason, the metric system is also known as the international system, which in other languages is referenced as the System International, or SI system. The Metric System is easier to use because everything is based on powers of 10. The following table provides the base values in the metric system: Prefix Symbol Power of 10 Decimal Equivalent Tera T 12 1,000,000,000,000 Giga G 9 1,000,000,000 Mega M 6 1,000,000 Kilo k 3 1,000 Hecto h Deka da 1 10 Standard Unit Deci d Centi c Milli m Micro u Nano n Pico p The standard unit is dictated by the type of measurement that is being done according to the following table: Measurement Unit Symbol Length Meter m Mass Gram g Volume (liquid) Liter L Current Ampere (amp) A Power Watt W Resistance Ohm Ω Time Second s Temperature Celsius C Using these charts, measurements can be converted from one prefix to another as in the following examples. 21

22 Example 1 Convert 10 km to dm. Solution: Looking at the charts, it can be determined that km is the abbreviation for kilometers (k = kilo and m = meters). Similarly, dm is the abbreviation for decimeters The exponent for km is 3, and the exponent for dm is 1 The exponents are subtracted to find how many places the decimal point needs to be moved: 3 (-1) = 4 The decimal point is moved 4 places. Since decimeters are smaller than kilometers, it is moved to the right 10 km = 100,000 dm Example 2 Convert hm to m. Solution: Looking at the charts, it can be determined that hm is the abbreviation for hectometers and m is the abbreviation for meters (since it has no prefix, it is just the standard unit) The exponent for hm is 2, and the exponent for m is 0 (the standard unit always has an exponent of 0) The exponents are subtracted to find how many places the decimal point needs to be moved: 2 0 = 2 The decimal point is moved 2 places. Since meters are smaller than hectometers, it is moved to the right hm = 0.32 m 22

23 Example 3 Convert 432,000 cm to dam. Solution: Looking at the charts, it can be determined that cm is the abbreviation for centimeters and dam is the abbreviation for dekameters The exponent for dam is 1, and the exponent for cm is 2 The exponents are subtracted to find how many places the decimal point needs to be moved: 1 (-2) = 3 The decimal point is moved 3 places. Since dekameters are larger than centimeters, it is moved to the left 432,000 cm = 432 dam One of the powerful aspects of the metric system is that the standard unit does not affect the way the conversions are done. In the following examples, notice how the standard unit changes, but the process is the same as for the above examples in meters (and would be the same for watts, liters, seconds, etc.). Example 4 Convert A to ma. Solution: Looking at the charts, it can be determined that A is the abbreviation for Amp and ma is the abbreviation for milliamps The exponent for A is 0, and the exponent for ma is 3 The exponents are subtracted to find how many places the decimal point needs to be moved: 0 (-3) = 3 The decimal point is moved 3 places. Since ma are smaller than A, it is moved to the right A = 1.3 ma 23

24 Example 5 Convert 1,000,000,000,000 ug to kg. Solution: Looking at the charts, it can be determined that ug is the abbreviation for microgram and kg is the abbreviation for kilogram The exponent for ug is 6, and the exponent for k is 3 The exponents are subtracted to find how many places the decimal point needs to be moved: 3 (-6) = 9 The decimal point is moved 9 places. Since kg are larger than ug, it is moved to the left 1,000,000,000,000 ug = 1,000 kg 24

25 2.2 Areas and Volumes Care must be taken when areas or volumes are involved. The following example demonstrates how to convert square cm (cm 2 ) to square dam (dam 2 ). Again, the standard unit itself (i.e. m 2 ) is inconsequential to the process. Example 1 Convert 25,000 cm 2 to dam 2. Solution: Looking at the charts, it can be determined that cm 2 is the abbreviation for square centimeters, and dam 2 is the abbreviation for square dekameters The exponent for cm is 2, and the exponent for dam is 1 The exponents are subtracted and then multiplied by 2 because this is an area measurement to find how many places the decimal point needs to be moved: 1 (-2) = 3, and then 3 x 2 = 6 The decimal point is moved 6 places. Since dam 2 are larger than cm 2, it is moved to the left 25,000 cm 2 = dam 2 25

26 Similarly, volumes will use a factor of 3. The following demonstrates a volume conversion. Example 2 Convert 3.8 Mm 3 to m 3. Solution: Looking at the charts, it can be determined that Mm 3 is the abbreviation for cubic megameters, and m 3 is the abbreviation for cubic meters The exponent for Mm is 6, and the exponent for m is 0 The exponents are subtracted and then multiplied by 3 because this is a volume measurement to find how many places the decimal point needs to be moved: 6 (0) = 6, and then 6 x 3 = 18 The decimal point is moved 18 places. Since m 3 are smaller than Mm 3, it is moved to the right 3.8 Mm 3 = 3,800,000,000,000,000,000 m 3 or 3.8 x m 3 26

27 2.3 Temperature In the metric system, temperatures are handled in a completely different way than other measurements. The Celsius scale provides the unit of measurement (you may hear about the Kelvin scale which is also used in the metric system but it is used primarily for more scientific and engineering applications). In the English system, the Fahrenheit system is used. The following table represents 3 important temperatures on the scales: Celsius temperature Fahrenheit temperature Significance of measurement Boiling point of water 0 32 Freezing point of water Absolute zero The following formulas provide the means to convert between these two temperature systems: C ( F F 5 9 C ) 32 Example 1 Convert 22.3 C to F. Solution: Since we are converting to Fahrenheit, we will use the second formula listed above 9 F 5 C32 F 9 (22.3) 32 5 F F So 22.3 C = F 27

28 Example 2 Convert F to C. Solution: Since we are converting to Celsius, we will use the first formula listed above C ( F 32) C ( ) C ( 45.67) C So F = C 28

29 2.4 Converting between systems It is often necessary to convert from one number system to the other. Because the systems are so different and have so many units to deal with, many times some research must be done to determine how one unit relates to another. Example 1 Chicago, IL and New York City, NY are approximately 800 miles apart. How far is this distance in km? Solution: A quick internet search reveals that 1 mi = km. In order to convert mi to km, this conversion factor must be used as follows: km 800mi 800(1.609 ) km mi Example 2 What is the weight capacity in kg of a 0.5 ton truck? Solution: A quick internet search reveals that 1 ton = kg. In order to convert tons to kg, this conversion factor must be used as follows: kg 0.5tons 0.5( ) kg ton 29

30 Chapter 2 Homework problems Section 2.1 Homework 1. Convert 46,120 cm to: a. mm b. m c. dam d. km 2. Convert 17 MW to: a. mw b. W c. kw d. GW 3. Convert L to: a. dl b. cl c. kl d. dal Section 2.2 Homework 1. Convert 1,000,000 mm 2 to: a. um 2 b. m 2 c. km 2 2. Convert a. cm 3 b. m 3 c. Tm 3 km 3 to: 30

31 Section 2.3 Homework 1. Fill in the following chart: Degrees Fahrenheit Degrees Celsius A third temperature scale that is often used in Science and Engineering applications is the Kelvin scale. The units on the Kelvin scale are the same as those on Celsius, except the scale is shifted so that zero on the Kelvin scale is the coldest possible temperature (known as absolute zero). The following equation relates Kelvin and Celsius: K C 273. Use this formula to verify the values of absolute zero on the Celsius and Fahrenheit scales that were given earlier in the text. Section 2.4 Homework (an internet search may be required to obtain the necessary conversion factors). 1. Convert 7.2 gallons to liters. 2. Convert 10,000 feet to meters. 3. Convert 727 grams to pounds. 4. Convert 10 hectares to acres. 5. Convert 2800 cm 3 to in 3. 31

32 Chapter 3 Percentages, precision, error, and tolerance intervals 3.1 Percentages: basic conversions Percentages are another way to numbers. If one half of a quantity is being discussed, that quantity can be represented as a fraction or a decimal as follows: Using a percentage, the decimal representation is multiplied by 100. So a third way to represent this quantity is illustrated as follows: % 2 In general, conversions are done according to the rules below: To convert a decimal to a percentage, move the decimal point 2 places to the right To convert a percentage to a decimal, move the decimal point 2 places to the left To convert a fraction to a percentage, first convert it to a decimal and then move the decimal point 2 places to the right The following table provides examples of these relationships: Fractional representation Decimal representation Percentage representation % % % % 32

33 Example 1 Convert 0.31 to a percentage Solution: To convert a decimal to a percentage, the decimal point is moved 2 places to the right: 0.31 = 31% Example 2 Convert 43.21% to a decimal Solution: To convert a percentage to a decimal, the decimal point is moved 2 places to the left: 43.21% = Example 3 Convert Solution: 4 11 to a percentage To convert a fraction to a percent, the fraction is converted to a decimal and then the decimal is converted to a percent: So % % 33

34 3.2 Percentages: base, part, and rate There are times when what percentage one number is of another needs to be determined. For example, 700 is 50% of In this case, we have the following: 1400 is called the base (B in the formulas below) 700 is called the part (P in the formulas below) 50% is called the rate (R in the formulas below) Note: the decimal version of R is used in the formulas below The formula that relates these components together are: P BR used when the part needs to be found P B used when the base needs to be found R P R used when the rate needs to be found B Example 1 What is 38% of 654? Solution: Here 38% is the rate (0.38 is used in the formula), and 654 is the base. Since we need to find the part, the problem is solved as follows: P BR P P So is 38% of

35 Example 2 29 is what percent of 514? Solution: Here 29 is the part and 514 is the base. Since we need to find the rate, the problem is solved as follows: P R B 29 R 514 R So 29 is 5.64% of 514 Example 3 63 is 0.03% of what number? Solution: Here 63 is the part and 0.03% is the rate. Since we need to find the base, the problem is solved as follows: P B R 63 B B 210, 000 So 63 is 0.03% of 210,000 35

36 3.3 Precision and error The precision of a measurement is defined to be the place of the digit that the measurement was rounded to. The following table represents the precisions of sample measurements: Measurement Precision Explanation 170 m 10 m Measurement was rounded to the tens place 420,000 mi 10,000 mi Measurement was rounded to the ten thousands places A A Measurement was rounded to the thousandths place 0.10 ft 0.01 ft Measurement was rounded to the hundredths place (as indicated by the last zero) Measurement was rounded to the ones place (as indicated by the tagged zero) 4250 cm 1 cm The precision of a measurement allows for different analytical values to be calculated related to error. The following definitions are associated with this concept: Greatest possible error one half of the precision Relative error greatest possible error divided by the actual measurement Percent of error relative error expressed as a percent Example 1 Find the precision, greatest possible error, relative error, and percent of error in the measurement 7200 m. Solution: The precision is 100 m since the measurement was rounded to the hundreds place The greatest possible error is 50 m since that is one half of the precision The relative error is since The percent of error is 0.694% since that is the percentage representation of the relative error 36

37 Example 2 Find the precision, greatest possible error, relative error, and percent of error in the measurement A. Solution: The precision is A since the measurement was rounded to the ten-thousandths place The greatest possible error is A since that is one half of the precision The relative error is since The percent of error is 4.17% since that is the percentage representation of the relative error 37

38 3.4 Tolerance intervals When manufacturing items that are to meet a specific measurement, no process can be perfect due to what is called common cause variation. As a result, manufacturers must be presented with tolerance intervals that indicate how much a particular measurement can differ from the specifications. The following definitions are associated with this concept: Tolerance acceptable amount a measurement may vary from the specification Upper tolerance limit the largest measurement that is still within tolerance Lower tolerance limit the smallest measurement that is still within tolerance Tolerance interval twice the tolerance The following table represents sample tolerance values: Specification Tolerance Upper limit Lower limit Tolerance Interval m m m m m 60 g 0.05 g g g 0.1 g 110 lb 2 lb 112 lb 108 lb 4 lb Example 1 If an order is placed for a widget with a length specification of 2.12 mm and a tolerance of mm, find the upper limit, lower limit, and tolerance interval. Solution: The upper limit is = mm The lower limit is = mm The tolerance interval is 2 x =

39 Example 2 If an order is placed for a widget with a weight of 170 kg and a tolerance interval of 15, find the tolerance, the upper limit, and the lower limit. Solution: The tolerance is 15/2 = 7.5 kg The upper limit is = kg The lower limit is = kg 39

40 Chapter 3 Homework problems Section 3.1 Homework Fill in the blank spaces in the following table: Fractional representation Decimal representation Percentage representation % 17.33% % Section 3.2 Homework 1. What is 14% of 150? 2. What is 11.35% of 720? 3. What is 0.001% of 4410? is what percent of 22? 40

41 is what percent of 100? is what percent of 96? is 32% of what number? is 0.37% of what number? is 300% of what number? 10. If Julie received a 7% raise and her new salary is $33,560, what was her original salary? Section 3.3 Homework Fill in the blank spaces in the following table: Measurement Precision Greatest Possible Error Relative Error Percent of Error 30 V 21.5 m g 105,000,000 mi Section 3.4 Homework Fill in the blank spaces in the following table: Specification Tolerance Upper limit Lower limit Tolerance Interval 17 m 1 m ml ml 0.5 oz 1.5 oz W cm cm 4.5 m 1 m 41

42 Chapter 4 Number systems 4.1 Decimal number system The typical system that is used in everyday mathematics is the decimal number system, also known as the base-10 system. The system is structured based on powers of 10 (hence the name of base-10). For example, the number 50,413 can be written in expanded form as follows: 50, 413 5(10000) 0(1000) 4(100) 1(10) 3(1) or more specifically: , 413 5(10 ) 0(10 ) 4(10 ) 1(10 ) 3(10 ) The place of every digit is important, as it indicates the power of ten that number is associated with. Another aspect of the base-10 number system is that there are 10 digits necessary to represent numbers in the system. These are the familiar digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. Every number in base-10 can be represented by using these 10 digits and the concept regarding their place in the number. Furthermore, all of the familiar operations (+,, x, ) can be performed on these numbers. Since electronic systems (such as computers) only store data in terms of on and off, they are unable to directly store a 10-state system. This concept will be discussed in the following sections. 42

43 4.2 Electronic machines Computers, calculators, and other electronic devices are sometimes known as finite state machines. This references the fact that these devices have memory which can be in any of a number of different states. More importantly to this discussion, because these machines are electronic they ultimately can only store on and off sequences. Because there are only two different options, machines do not have the ability to store a 10-digit sequence like the numbers in a base-10 system. To simulate this system, we need to use a binary number system (or base-2) in place of the more familiar base-10 system. However, different number systems are all structured the same way. The only thing that changes is the base that is used for each place where the digit resides. One example of how this manifests itself in technology is how a computer stores a symbol such as a letter A. Ultimately, this letter must be stored as a sequence of ons and offs so each letter (and in fact all letters, digits, symbols, etc.) is given a numeric code. Most personal computers use the Unicode system to encode these characters. In Unicode, the letter A has a code of 65. However, even the number 65 is one step removed from what the machine can store since 65 is in the decimal (10-state) base. 65 is converted to binary: 65 = in base 2 (groupings on computers are in multiples of 8). Hence, the number 65 actually looks like this sequence: off, on, off, off, off, off, off, on Numbers can be written in expanded form using the same technique as that which was used for base-10 as follows: ) 1(2 ) 0(2 ) 2 0(2 ) 1(2 ) 0(2 ) 1(2 ) 0(2 ) 1(2 Note the subscript on the number indicates that it is in base-2 as opposed to base-10. Furthermore, if you add the values on the right side of the equation, you get the original value of 65. While any positive integer can be used as a base, the bases relevant to a technological discussion are the following: Base Name Valid digits 2 Binary 0 and 1 8 Octal 0, 1, 2, 3, 4, 5, 6, and 7 10 Decimal 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 16 Hexadecimal (Hex) 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, and F Binary numbers are important because they provide a means of simulating the on/off aspect of electronic devices. Hexadecimal numbers are used in different technological applications such as networking and design (among others). Since octal numbers are no longer as prevalent as they used to be, they are not covered in this text. 43

44 4.3 Base-2 (Binary) As mentioned above, the binary number system has only 2 valid digits: 0 and 1. Furthermore, the place of each digit in a number indicates the associated power of 2. Any number in any non-decimal number base can be converted to base 10 by writing the number in expanded form and then calculating the products and sums on the right-hand side. Example 1 Convert to decimal Solution: is a binary number as indicated by the subscript. Writing this number in expanded form yields the following: (2 ) 0(2 ) 1(2 ) 1(2 ) 0(2 ) (2 ) 0(2 ) 1(2 ) 1(2 ) 0(2 ) 22 So the binary number is equivalent to the decimal number = 22 Example 2 Convert to decimal Solution: (2 ) 1(2 ) 1(2 ) 1(2 ) (Note the zeroes were ignored) (2 ) 1(2 ) 1(2 ) 1(2 ) 114 So =

45 When converting numbers from base-10 to base-2, a different approach is used as follows: Example 3 Convert 152 to binary Solution: represents the quotient and 0 represents the remainder when 153 is divided by The process stops when a 0 is obtained in the bracket Reading the remainders from the bottom to the top gives the result. 152 = Example 4 Convert 37 to binary Solution: =

46 4.4 Base 16 (Hexadecimal) Because the hexadecimal number system (often abbreviated to hex ) is base-16, it needs 16 valid digits. This is problematic because there are only 10 standard digits (0 through 9). Letters are commonly used for the remaining 6 digits according to the following chart: Digit in Hex Base-10 value 0 9 Standard meaning A 10 B 11 C 12 D 13 E 14 F 15 As in the other bases, the place of each digit is important. In hex, the placement indicates the associated power of 16. Converting base-16 numbers to decimal can be done by putting the number in expanded form and then calculating the products and sums on the right-hand side. Example 1 Convert 3CD2 16 to decimal Solution: 3CD2 16 is a hexadecimal number as indicated by the subscript. Writing this number in expanded form yields the following: CD2 3(16 ) 12(16 ) 13(16 ) 2(16 ) 15, So 3CD2 16 = 15,570 46

47 Example 2 Convert A141E2 16 to decimal Solution: A141E2 16 is a hexadecimal number as indicated by the subscript. Writing this number in expanded form yields the following: A E (16 ) 1(16 ) 4(16 ) 1(16 ) 14(16 ) 2(16 ) 10,568, So A141E2 16 = 10,568,162 When converting numbers from base-10 to base-16, the same division/remainder approach is used as that which was used for binary: Example 3 Convert 5038 to hexadecimal Solution: (which is E in base-16) (which is A in base-16) = 13AE 16 47

48 Example 4 Convert to hexadecimal Solution: (which is C in base-16) (which is D in base-16) = 32DC 16 Base-16 numbers can be converted to base-2 numbers using the following process: 1. Convert the base-16 number to base Convert the base-10 number to base-2 However, base-16 and base-2 numbers have a unique relationship whereby every 4 digits in a base-2 number equates to 1 digit in a base-16 number. This can be useful to provide a shortcut to converting numbers between these two bases. Example 5 Convert to hexadecimal. Solution: The digits in is broken up into groupings of 4: The groupings are now associated with their base-16 number using expanded form: = 1(2 3 ) + 1(2 2 ) + 1(2 0 ) = 13 = D = 1(2 0 ) = = 1(2 2 ) + 1(2 1 ) = 6 16 Placing the values in their respective locations gives = D

49 Example 6 Convert to hexadecimal. Solution: The digits in is broken up into groupings of 4 and replaced with their hex equivalents: D 5 So = 18D5 16 The reverse process is used to convert base-16 to base-2. Example 7 Convert A3 16 to binary. Solution: The digits in A3 are associated with their corresponding groupings of 4: A So A3 16 = Example 8 Convert 1CC 16 to binary. Solution: The digits in 1CC are associated with their corresponding groupings of 4: 1 C C So 1CC 16 =

50 Number Bases Chart The integers 0 32 in Base 10, 2, 8, and 16 Decimal (Base 10) Binary (Base 2) Octal (Base 8) Hexadecimal (Base 16) A B C D E F A B C D E F

51 Chapter 4 Homework problems Section 4.2 Homework 1. Write 6192 in expanded form (the absence of a subscript indicates this is a base 10 number). 2. Write in expanded form 3. Write in expanded form 4. Write in expanded form 5. Write AE3 16 in expanded form 6. Write 13C2D0 16 in expanded form Section 4.3 Homework 1. Convert to Base Convert to Base Convert to Base Convert 41 to Base 2 5. Convert 547 to binary 6. Convert 1024 to binary Section 4.4 Homework 1. Convert AE3 16 to Base Convert 13C2D0 16 to Base Convert 712 to Base Convert 501,322 to Base 16 51

52 5. Convert to Base Convert to Base Convert A3 16 to Base 2 8. Convert 19CD 16 to Base 2 52

53 Chapter 5 Formulas and data substitution 5.1 Substituting data into formulas Many formulas are encountered in technical settings to show relationships among the values in a given situation. A common simple formula in electronics is Ohm s law which states: E IR where E is the voltage (measured in volts), I is the current (measured in amps) and R is the resistance (measured in ohms). Hence, the voltage in a circuit is equal to the current multiplied by the resistance. Example 1a Find the voltage in a circuit if the current is A and the resistance is 10,000Ω. Solution: E IR E A10,000 E 9V So the voltage in this circuit is 9V. Example 1b It is not uncommon for current and resistance values to be given in ua (microamps) and kω (kiloohms or k-ohms). As a result, it would be common to see the problem in Example 1a phrased this way: Find the voltage in a circuit if the current is 900uA and the resistance is 10kΩ. Solution: E IR E A10, 000 (since 900uA=0.0009A and 10k 10, 000 ) E 9V Using different (but equivalent) units does not change the situation, and hence the voltage is still 9V. 53

54 There are many formulas which will be encountered in technical settings. A small subset is provided here. Formula Description force = (pressure)(area) power = (current)(voltage) power = (square of the current)(resistance) Electric charge current= time 2 E b SN Engine displacement = (square of the bore)(stroke)(number of cylinders) 9 Fahrenheit temp = 9/5(Celsius temp) + 32 F C 32 5 f pa P IE P Q I t 2 I R Substituting data into formulas always uses essentially the same process. The values are substituted in and the calculations are performed to determine the unknown quantity. Example 2 Find the engine displacement given that the bore is 3.63 in, the stroke is 3.65 in, and the engine has 8 cylinders. Solution: E (3.63 )(3.65)(8) E 302in 3 Example 3 Water boils at 100 C. What is the equivalent Fahrenheit temperature? Solution: 9 F C32 5 F 9 (100) 32 5 F F 212 So water boils at 212 F. 54

55 5.2 Simple algebraic manipulation of formulas Problems encountered in technology are often not always in terms of the isolated variable. In one of the situations above, the formula was written in terms of V and the situation required that the value of V be found. However, it is quite possible that V would be a known quantity and one of the other two values would need to be found. In this case, minor algebraic manipulation would have to be performed to isolate the required variable. Example 1 A light bulb in a circuit has a voltage of 12V and current of 24mA. What is its resistance? Solution: E IR E R Both sides of the equation are divided by I I E R The equation is flipped to place R on the left I 12V R The known values are substitued in for E and I 0.024A R 500 The value of R is calculated (and left in since k is typically used for values over 1000) So the resistance is 500Ω. Example 2 What is the wattage of the bulb in example 1? Solution: P IE P (12 V )(0.024 A) The known values are substitued in for I and E P W The value for P is calculated and converted to mw P 288mW So the power is 288mW. 55

56 Example 3 As of the end of 2013, the highest recorded temperature in Normal, IL was F. What is this temperature in Celsius? Solution: 9 F C F 32 C 5 32 is subtracted from both sides of the equation 5 5 ( F 32) C both sides of the equation are multiplied by C ( F 32) 9 the equation is flipped to place C on the left 5 C ( ) 9 the known value is substituted for F C 42 the value of C is calculated So as of the end of 2013, the highest recorded temperature in Normal, IL was 42 C. 56

57 Chapter 5 Homework problems Section 5.1 Homework 1. The pressure between a boy and the ground he stands on is N/cm 2. If the boy s feet cover an area of 240 cm 2, what is his weight (in N)? 2. If the current in a circuit is 4A, and the voltage is 12V, what is the power? 3. If the current in a circuit is 2000mA, and the voltage is 9V, what is the power? 4. If the current in a circuit is 3A, and the resistance is 2 Ω, what is the power? 5. Find the engine displacement of a twin cylinder engine in cm 3 if the bore is 5.4 cm and the stroke is 4.0 cm. Section 5.2 Homework 1. If the Power in a circuit is 50W, and the voltage is 12V, what is the current? 2. If the Power in a circuit is 50W, and the current is 9A, what is the voltage? 3. If the Power in a circuit is 50W, and the current is 9A, what is the resistance? 4. If the Power in a circuit is 50W, and the resistance is 5 Ω, what is the current? 5. In the late 1960s, Toyota produced a small 1988 cm 3 6-cylinder engine. If the bore of the engine was 75 mm, what was the stroke? 57

58 Chapter 6 Basic Geometry 6.1 Angles Most shapes are in some way related to line segments and angles. There are many different relationships inherent in geometry, so it is important to first understand what angles are. An angle is formed when two line segments meet in a point. This point is known as the vertex of the angle. In the following image, O is the angle vertex: Angles can be measured using a few different approaches. One measurement systems uses degrees. Any angle that is formed by two perpendicular segments as in the image below is known as a right angle and has a measurement of 90. The question can reasonably be asked What is a degree? If a circle is cut into 360 equal pie-shaped wedges, each of these wedges is exactly 1, regardless of the size of the circle: Angles can now be classified as follows: An acute angle measures less than 90 A right angle measures exactly 90 An obtuse angle measures more than 90 Two angles are complements if the sum of their measures is 90 Two angles are supplements if the sum of their measures is

59 There are many angular relationships that exist when angles are formed by intersecting lines as follows. In the following image, angles A and B are called vertical angles and will always have the same measurement. Angles C and D are also vertical angles. A somewhat more complicated example occurs when two parallel lines are intersected by a third line (called a transversal). The image and table below show the various relationships that result in this scenario (the table does not show all possible relationships). Angle pair Relationship Reason A and D Equal Vertical angles D and E Equal Alternate-interior A and C Supplements Form a straight line C and E Supplements E = A and A and C are supplements E and G Supplements Form a straight line A and E Equal A = D = E A and G Supplements A = E and E and G are supplements 59

60 6.2 Triangles Polygons are closed figures that are made up of line segments. They are categorized by the number of sides. Triangles are the 3-sided polygons where the sum of the interior angles is always 180. There are two different ways to categorize triangles as follows: Categorization 1 based on the interior angles A triangle is acute if all three angles measure less than 90 A triangle is right if one of its angles measures 90 (the other two angles will always be less than 90 and will be complements of each other). The longest side of a right triangle is called the hypotenuse, and the two shorter sides are called the legs. A triangle is obtuse if one of the angles measures more than 90 (the other two angles will always be less than 90 ) Categorization 2 based on the lengths of the sides A triangle is scalene if all three sides have different lengths A triangle is isosceles if two of the sides have the same length A triangle is equilateral if all three sides have the same length Example 1 What is the measure of angle A in the image below? Solution Since A is a right triangle (indicated by the small square in the lower-left corner), the other two angles are complements of each other. The measure of the angle A can be found by subtracting the other angle from 90 : A = =

61 Example 2 An isosceles triangle has a perimeter of 22m and the shorter side is 4m. What is the length of the two equal sides? Solution The perimeter of a figure is the distance around the figure. In the case of a triangle, it is the sum of the 3 sides. Using L as a variable for the length of the equal sides gives the following: 4 + L + L = L = 22 2L = 18 L = 9 So the length of the two equal sides is 9m. The Pythagorean Theorem When the lengths of two sides of a right triangle are known, the third side can always be calculated using the Pythagorean Theorem. This theorem is demonstrated as follows: In the image below, c a b 61

62 Example 3 Find the length of the unknown side in the following figure: Solution This is a right triangle so the Pythagorean Theorem is used. Since x represents the hypotenuse, the solution is calculated as follows: x x x x 985 x 31.4 So the length of x is 31.4 cm. Care needs to be taken because the variable c here is always indicating the hypotenuse. The following problem indicates how to solve for one of the legs. 62

63 Example 4 Find the length of the unknown side in the following figure: Solution This is a right triangle so the Pythagorean Theorem is used. Since x represents one of the legs, the solution is calculated as follows: 100 x x x x x x 92.9 x So the length of x is 92.9 mm. 63

64 Area The formula for the area of a triangle is given by following example applies this formula. A 1 2 bh where b is the base and h is the height. The Example 5 Find the area of the triangle. Solution The height of this triangle is 19 in and the base is 42 in. A 1 2 A (42 in)(19 in) A 1 2 bh (798 in ) A 399in 2 So the area of the triangle is 399 in 2. 64

65 In a right triangle, the two legs represent the base and the height. Example 6 Find the area of the triangle. Solution The height of this triangle is 14.3 m and the base is 7.1 m. A 1 2 A (14.3 m)(7.1 m) A 1 2 bh ( m ) A m 2 So the area of the triangle is m 2. 65

66 Example 7 Find the area of the triangle. Solution Since only the length of one of the legs is known, the other must be found using the Pythagorean Theorem x x x x x x cm x Now that the 2 legs are known to be 20.3 cm and cm, the area can be found. A 1 2 A (20.3 cm)(18.55 cm) A 1 2 bh ( cm ) A 188.3cm 2 So the area of the triangle is 188.3cm 2. 66

67 6.3 Quadrilaterals Quadrilaterals are the four-sided polygons. Two of the common types found in a technical setting are parallelograms and trapezoids. Parallelograms have two pairs of parallel sides. Examples are rectangle and squares, although not all parallelograms have 4 right angles. Trapezoids have one pair of parallel sides. Name Area formula Example General parallelogram A bh Rectangle A lw Trapezoid A 1 a b h 2 ( ) Example 1 Find the area of the parallelogram. Solution The base of the parallelogram is 9.5 in and the height is 7 in. A bh A (7 in)(9.5in) A 66.5in 2 So the area of the parallelogram is 66.5 in 2. 67

68 Example 2 Find the area of the trapezoid. Solution The height of the trapezoid is 5.1 ft and the two bases are 13.2 and 16.8 ft. A ( a b) h A A ( )(5.1) (30)(5.1) A 76.5 ft 2 So the area of the trapezoid is 76.5 ft 2. 68

69 6.4 Circles Circle terminology and basic formulas Important terminology associated with circles is discussed in the following table. The most important of these are the radius and diameter. Line / Line segment Chord Diameter Radius Secant Tangent Description Line segment that begins on one point of the circle and ends on the other Chord that goes through the center of the circle Line segment starting at the center of the circle and going to a point on the circle Line that goes through the circle, intersecting with it in two points Line that intersects with the circle in one point Two important formulas associated with circles given below. For this text, the value of will be used for the constant π. Formula Description C 2 r Circumference (distance around the circle) 2 A r Area Example 1 Find the circumference and area of the circle. Solution The radius of this circle is 6 cm. C 2 r C 2( )(6 cm) C 37.7cm A r 2 A ( )(6 cm) A 113.1cm 2 2 So the circumference of the circle is 37.7 cm and the area is cm 2. 69

70 Example 2 Find the diameter of a circle that has an area of 48.2m 2. Solution First the area formula is algebraically manipulated to isolate the radius, and then the values are substituted. Once the radius is found, it is doubled to find the diameter. A r A r 2 2 A r A r m r r 3.9m Since the radius of the circle is 3.9m, the diameter is 7.8m. Arcs and associate angles Arcs represent a portion of the circle and are measured in degrees using the same measurement system as for angles (i.e. 1/360 th of the way around a circle represents 1 ). 70

71 For example, in the following image, the red arc labeled s measures 90 since it is ¼ of the circle: Angles that originate from the center of a circle are called central angles. These angles have the same measure as the corresponding arc on the circle as demonstrated by the following image: Angles that originate from a point on a circle are called inscribed angles. These angles also have a relationship to the corresponding arc on the circle. The measure of an inscribed angle is always half the measurement of the corresponding arc as demonstrated by the following image: 71

72 A relationship involving triangles and circles is that a triangle inscribed in circle, and using the diameter as one of the sides of the triangle, is guaranteed to be a right triangle. Furthermore, the diameter of the circle will always represent the hypotenuse of the right triangle. Hence the red triangle in the image below (where C marks the center of the circle) must have a 90 angle as marked in the image. 72

73 Example 3 If C is the center of the circle, find length of the unknown side x in the triangle, and the measure of the arc s. Solution The red triangle must be a right angle (since it is inscribed in a circle and the diameter represents one of its sides). This information will be used to find the values. First the length of x can be found using the Pythagorean theorem. x x x 4.0m 2 2 Next, the measure of the arc s can be found using the following two pieces of information: 1. The angle opposite the hypotenuse is The sum of the angles in a triangle is The unknown angle is the inscribed angle associated with the arc s The third angle in the triangle is calculated as follows: Inscribe angle measure = = 59 And the corresponding arc s is calculated as follows: S = 59 x 2 = 118 Combining what was calculated above, the values are: x = 4.0 m s =

74 Chapter 6 Homework problems Section 6.1 Homework 1. If an angle measures 38, then: a. what is the angle s complement (if it exists)? b. what is the angle s supplement (if it exists)? 2. If an angle measures 114, then: a. what is the angle s complement (if it exists)? b. what is the angle s supplement (if it exists)? 3. In the image below, what are the two pairs of equal angles? 4. In the image below, find 3 pairs of equal angles and 3 pairs of supplementary angles. 74

75 Section 6.2 Homework 1. Fill in the chart for each triangle: Triangle Acute, right, or obtuse Scalene, Isosceles, or Equilateral 2. In the image below, find the measure of angle A: 3. In the image below, find the value of x and the perimeter of the triangle. 75

76 4. In the image below, find the value of x and the perimeter of the triangle. 5. Find the area of the triangle in question Find the area of the triangle in question Find the area of the following triangle: 8. If an isosceles triangle has a perimeter of 176 ft, and the shorter side is 10 feet less than the measure of the two equal sides, what is the length of the three sides of the triangle? 9. Find the area of the triangle in question 8. 76

77 10. When the lengths of the three sides of a triangle are known, but the height is not, Heron s formula can be used to calculate the area. Heron s formula states that if a, b, and c represent the three sides, then the area is calculated as follows: abc A s( s a)( s b)( s c) where s 2 Use this formula to find the area of the following triangle: Section 6.3 Homework 1. Find the area and perimeter of a rectangle with a length of 30m and a width of 42m. 2. Find the length of a rectangle with an area of 700cm 2 and a width of 68cm. 3. Find the area of the trapezoid in the image below: 4. Find the area and perimeter of the trapezoid in the image below: 77

78 5. A building has a length of 40 feet, a width of 60 feet, and a height of 80 feet. Find the amount of paint needed (in gallons) and the cost of the paint if: Section 6.4 Homework a. all 4 sides of the building needed to be painted b. a gallon of paint covers 420 square feet c. a gallon of paint costs $11 1. Fill in the values in the table below. Circle radius Circle diameter Circle circumference Circle area 3.4 in 17 ft 3 x 10 6 km 104 cm 712 mm 3 2. Find the measure of the arc s: 78

79 3. Find the measure of the arc s: 4. Find the measure of the arc s: 5. Find the measure of the arc s, and the diameter of the circle: 79

80 Chapter 7 Basic Trigonometry 7.1 The Trigonometric ratios The six trigonometric ratios are relationships between the lengths of the sides in a right triangle. They are defined in the following chart (in this chart, A represents an angle). Full ratio name Standard abbreviation Ratio sine(a) sin(a) side opposite A commonly referred to as hypotenuse O H cosine(a) cos(a) side adjacent to A commonly referred to as hypotenuse A H tangent(a) tan(a) side opposite A side adjacent to A commonly referred to as O A secant(a) sec(a) hypotenuse side adjacent to A commonly referred to as H A cosecant(a) csc(a) hypotenuse side opposite A commonly referred to as H O cotangent(a) cot(a) side adjacent to A commonly referred to as side opposite A A O Secant, cosecant, and cotangent can all be found if the other ratios are known. As a result, they do not appear on scientific calculators, they are not often used in technical classes, and they will not be covered in this text. The acronym SOHCAHTOA (pronounced soh-ca-toe-a) is often used to remember the ratios of the three O A O major trigonometric functions, sincesin A, cos A, and tan A. H H A 80

81 Example 1 Find sina, cosa, and tana in the following triangle. Solution O 13 sin A H 21 A 17 cos A H 21 O 13 tan A A 17 81

82 Example 2 Find sinb, cosb, and tanb in the following triangle. Solution Since the hypotenuse is needed to find sin and cos, that is calculated first using the Pythagorean Theorem: h The three trigonometric ratios can now be calculated: O 8.2 sin B H 9.5 A 4.7 cos B H 9.5 O 8.2 tan B A 4.7 m 82

83 7.2 Finding trigonometric values on a calculator All scientific calculators can calculate any of the trigonometric values of a particular angle (and can also calculate an angle given the trigonometric value). Angles can be measured using different measurement systems. To be consistent with the previous angular discussions, the degree measurement system will be used. When using a scientific calculator, care must be taken to ensure that the degree measurement system is used (as opposed to radians or gradians), as this will affect the answers. In this text, the following two assumptions are made: 1. The calculator used is the standard Windows OS calculator in Scientific mode (other calculators can be used, but they all have their own processes) 2. The degree measurement system is used This image shows what the calculator looks like. Note the Degrees radio button is selected. The sin, cos, and tan buttons are also visible in this mode. 83

84 Using the calculator to determine trigonometric values of angles. Example 1 Find cos(58 ). Solution The question is asking for the cosine of a 58 angle to be found. This can be done by entering 58 into the calculator and pressing the cos button. The calculator displays the value (trig values are often rounded to four significant digits), and so: cos(58 ) = Example 2 Find the sin, cos, and tan of Solution Using the calculator, the following values are obtained: sin(71.3 ) = cos(71.3 ) = tan(71.3 ) =

85 Using the calculator to determine angles given their trigonometric values. Example 3 If cosa = , find the measure of angle A. Solution The question is the opposite of what is being asked in example 1 above. Rather than being given the angle and being asked for the trig value, this question gives the trig value and is asking for the angle. Scientific calculators can calculate this, but the process is slightly different. To calculate this value, is entered in the calculator, and the Inv button is pressed (this is often marked as the 2 nd key on Scientific calculators). The image below shows the calculator at this point (the Inv button is in yellow): The cos -1 button replaces the cos button on the calculator. Pressing this button yields the value 71.69, and so: If cosa = , then the measure of angle A =

86 Example 4 If sina = , find the measure of angle A. Solution Using the calculator, the following value is obtained: If sina = , then the measure of angle A = Example 5 If tana = 2.668, find sina and cosa. Solution To find sina and cosa, first the measure of A must be found. Using the calculator (as in examples 3 and 4), the following is obtained: If tana = 2.668, then the measure of angle A = Now, cosa and sina can be found (as in examples 1 and 2): sin(69.45 ) = cos(69.45 ) =

87 7.3 Using trigonometry to find unknown components of a triangle The concepts in sections 7.1 and 7.2 can be used to find the lengths of sides and measures of angles in right triangles if enough information is provided. Example 1 Find x in the image below. Solution: In this example, the 61 angle and its opposite side are provided. The adjacent side must be found. Since the sides in question are the opposite and adjacent sides, the appropriate O trigonometric function to use is tangent (since tan A ). A The calculation is performed as follows: 14.1 tan 61 x x(tan 61) 14.1 x 14.1 tan 61 x And so x = m. 87

88 There are many applications of right triangles in technical courses. The following provides one such example. Example 2 The relationship between Resistance (R), Inductive reactance (X L), and Impedance (Z) (all three of these are measured in Ohms), and the phase angle in a circuit is given by the image below. If R = 11Ω and Z = 15Ω, find X L and the phase angle ( ). Solution: X L can be found by using the Pythagorean Theorm: 2 2 X L can be found using R, Z, and cos( ). Cos is used because R and Z represent the adjacent angle and the hypotenuse. R 11 cos Z 15 Since cos , using cos on the calculator yields the following: So: X L = 10.2Ω and =

89 Chapter 7 Homework problems Section 7.1 Homework 1. Find sina, cosa, and tana. 2. Find sina, cosa, and tana. 3. Find sinb, cosb, and tanb. Section 7.2 Homework 1. If A = 67, find: a. sina b. cosa c. tana 2. if B = 23, find: a. sina b. cosa c. tana 89

90 3. Was there any relationship between the sin and cos values in question 1 and 2? If so, explain why this happened. 4. If sina = , find A. 5. If cosa = , find A. 6. If tana = 3.614, find A. 7. Explain what happens (and why) if you try to find A if cosa = If sina = , find cosa and tana. Section 7.3 Homework 1. Find x and y in the image below. 2. Find x and y in the image below. 90

91 3. Find the measure of angle A in the image below. 4. Find the measure of angle A in the image below. 5. Find the measure of angle A in the image below. 91

92 For questions 6, 7, and 8, use the image below that represents the relationships between Impedance, Reactance, Resistance and the phase angle in a circuit: 6. If R and X L are both 14Ω, find Z and the phase angle. 7. If Z = 20Ω and = 38, find R and X L. 8. If R = 8.5Ω and Z = 12Ω, find and X L. 92

93 Chapter 8 Interpreting Charts and Graphs 8.1 Charts and Graphs Charts and graphs provide a visual representation of data. The data can be representative of many different situations, from quality control processes to informational trends over time to the power output of a device. While charts and graphs present the data in a concise and more easily-read format, it takes practice to become comfortable reading and analyzing these charts. Furthermore, there are many different kinds of charts that can be used. Many software products provide ways to enter data and build charts and graphs. All of the images in this section were either built from Excel, or by using the TI-83 graphing calculator. Three samples are given below, and these three types are discussed in more detail on the following pages. Chart/Graph type Bar chart Sample Image Series2 1, 23% Pie chart 3, 47% 2, 30% Coordinate system graph Bar charts As the name indicates, bar charts use bars to present information. The bars in the chart can be presented either horizontally or vertically. 93

94 Defective parts Technical Mathematics Example 1 The following chart shows the number of defective parts produced during a manufacturing process. Quality control made a change to the process in September to reduce the number of defective parts. Quality control report - Defective Parts in Process Jan & Feb Mar & Apr May & Jun Jul & Aug Sep & Oct Nov & Dec Timeframe Chart analysis 1) Question: How many parts were defective in January and February? Answer: Looking at the first blue bar, it appears that approximately 720 parts were defective in these months. 2) Question: How many parts were defective in the months January through August? Answer: These months comprise the first 4 bars. Adding them together, it appears that there were approximately = 2915 defective parts in this timeframe. 3) Question: Did the change made to the process in September seem to have a positive affect? Answer: Additional analysis needs to be done to answer this question. A good approach would be to calculate the average number of defective parts per month. For Jan Aug, the average is: defective parts per month. 8 For Sep Dec, the average is: defective parts per month. 4 While more data and analysis should be done to verify the findings, it appears that the process did have a positive effect since the average number of defective parts per month was lowered by

95 Employee Technical Mathematics Example 2 The following chart is a clustered (or stacked) horizontal bar chart showing information regarding employee salaries and taxes. Payroll Information Lori Williams Joe Black John James Lisa Johnson Jake Smith 0 20,000 40,000 60,000 80, , ,000 Dollar Value Net pay Taxes Gross salary Chart analysis 1) Question: Who paid the most in taxes? Answer: Looking at the legend at the bottom of the chart show that the orange bars represent taxes. Lori Williams has the longest orange bar, paying approximately $16,000 in taxes. 2) Question: What is the average employee net salary? Answer: The green bars indicate the net salary. The average is calculated as follows: 88, , , , , , ) Question: What is the total company payroll? Answer: To find the total payroll, the gross salaries needed to be added. Payroll = 110, 00 41, , , , ,

96 Pie charts As the name indicates, pie charts use circles with wedges. The wedges are reminiscent of pie-shaped pieces giving them their name. Example 3 The following chart shows the same information as in Example 1, only this time a pie-chart is used. Since this is simply a different way of presenting the data, all of the same questions can be answered. Defective parts (total 4004 defective parts) Nov & Dec 14% Jan & Feb 18% Sep & Oct 13% Mar & Apr 18% Jul & Aug 17% May & Jun 20% Jan & Feb Mar & Apr May & Jun Jul & Aug Sep & Oct Nov & Dec Chart analysis 1) Question: How many parts were defective in January and February? Answer: Looking at the blue area, 4004*.18 = 720 parts were defective in this timeframe. 2) Question: How many parts were defective in the months January through August? Answer: Looking at the blue, red, green, and purple areas, 4004 *.73 = 2923 parts were defective in this timeframe. 3) Question: Did the change made to the process in September seem to have a positive affect? Answer: For Jan Aug, the average percent is % 8 For Sep Dec, the average is: %, so again, it seemed to have a positive effect. 4 96

97 Example 4 The following chart is a version of a pie chart that focuses on a specific part of the chart and provides a secondary pie chart with only that data. May & Jun 20% Defective parts (total 4004 defective parts) Jul & Aug 17% Nov & Dec 14% Other 27% Mar & Apr 18% Sep & Oct 13% Jan & Feb 18% Jan & Feb Mar & Apr May & Jun Jul & Aug Sep & Oct Nov & Dec Chart analysis This chart not only shows the defective part totals broken out in their months, but also shows the timeframe associated with the manufacturing process change. This allows for more detailed analysis and easier readability so that appropriate decisions can be made regarding the effectiveness of the change, and if that approach should be continued. Coordinate system graphs Coordinate system graphs use a standard x and y axis approach. The x-axis is the horizontal axis and the y-axis runs vertically. Sometimes these axes will indicate other values (such as time, power, dollar values, or any value that can be measured). 97

98 Example 5 The following graph displays the relationship from Ohm s law discussed earlier in the text. Recall. For a set value of R = 5Ω, the graph appears in the image below. Note that E is the vertical axis and I is the horizontal axis. that Ohm s law states E IR Chart analysis In interpreting this chart, it can be seen that there is a smooth linear relationship between E and I. For every unit I goes up, E goes up 5 units. This can be seen in the two images below: When I = 1, E = 5: When I = 2, E = 10: 98

99 Example 6 2 The following graph displays the relationship Power relationship P I R (given earlier in the text). For a set value of R = 5Ω, the graph appears in the image below. Note that P is the vertical axis and I is the horizontal axis. Chart analysis In interpreting this chart, it can be seen that the relationship is no longer linear. Unlike in the first example, this time as the value of I increases the value of P increases dramatically. This can be seen in the two images below: When I = 1A, P = 5W: When I = 2A, P jumps all the way to 20W (note that the point is off the graph below). This increase will get more and more dramatic as I gets larger due to the exponential nature of the formula. 99

100 Wins Technical Mathematics Chapter 8 Homework problems Section 8.1 Homework 1. Use the chart below to answer the questions that follow Chicago Cubs win totals per season Season a. In what year did the Cubs win the most games? b. Approximate the total wins for all even numbered years. c. If a season consists of 162 games, approximate the number of losses in the year d. How many games have the Cubs won since the 2005 season (include 2005 in your total)? e. How many games have the Cubs lost since the 2005 season (include 2005 in your total)? 100

101 2. A small company reported $70,000 of revenue during the calendar year. The chart below indicates the percentage of that revenue that occurred during each quarter of the year. Use the chart to answer the questions that follow. Revenue 33% 21% 20% 26% Q1 Q2 Q3 Q4 a. What percent of the revenue was generated during the first half of the year? b. How much combined revenue (in dollars) was generated in the months of July, August, and September? c. How much combined revenue (in dollars) was generated in the months of April, May, June, October, November, and December? d. A fiscal report provided by the company claims that their sales are greatest in the months leading up to the holiday season. Explain whether or not the chart supports this claim. 101

102 3. The chart below uses what is called a line graph to show the number of bolts produced by a company every 5 years. Use the chart to answer the questions that follow. Analysing bolt production 102, ,000 98,000 96,000 94,000 92,000 90,000 88,000 86, Bolts meeting specs Total bolts produced a. Approximately how many bolts were produced in 1990? b. Approximately how many bolts produced in 1990 did not meet specifications? c. Approximately what percent of bolts produced in 1995 did not meet specifications? d. A report was published by the company for its stock holders in The report stated that a new process would be put into production in 1995 and the expectation was that while there would be some inefficiencies when the process was first implemented, over time the process should allow for more bolts to be produced. While the chart only supplies a small subset of the data, does it corroborate the claims made by the company? e. How many bolts were produced in 1985? 102

103 4. The following two graphs show the relationship between Power, Current, and Resistance in 10 situations using the formula P 2 I R The first graph uses a standard scale: P=I 2 R Resistance Current (A) Power (W) Because there is an exponential relationship, a standard scale makes it difficult to see the smaller values of I and R (because P becomes large very quickly). In this case, a logarithmic scale is often used as in the following chart: P=I 2 R Resistance Current (A) Power (W) Use these charts to answer the question below: 103

104 a. In second chart, what is the Resistance in all 10 cases? b. In the first chart, can the Resistance be identified? If not, why? c. How does the current change from one scenario to the next? Which chart helps to answer this question? d. If the current increases from 10A to 20A, what is the change in the Power? e. If the current increases from 10A to 100A, what is the change in the Power? 5. Use the graph below to answer the questions that follow. a. For each value of x, find the corresponding value of y: i. X = 2 ii. X = 0 iii. X = 3 iv. X = 6 b. For each value of y, find all corresponding values of x: i. Y = 0 ii. Y = 2 iii. Y = 4 iv. Y = 3 104

105 Chapter 9 Basic Statistics 9.1 Data presentation Statistics in technical settings often deals with collecting and analyzing data. This process can manifest itself in many ways including but not limited to the following: Collecting historical data over time to allow for future predictions that can lead to better business or processing decision-making Collecting data to identify trends in industry Collecting data about a specific process that can lead to better quality control Collecting data about a specific process that can lead to the determination of whether the process can meet order specifications in terms of a customer s tolerance While there are different mathematical calculations and techniques involved in this process, first and foremost the data must be presented in a readable format. Consider the following chart of raw data: Measurement of widgets produced (target measurement: 2.00 mm) Widgets above target Widgets below target Looking at this data, it is hard to determine any kind of categorization or trend as it is not grouped together or in any kind of natural ordering. One common approach is to create what is known as a frequency table that will group and order the data in a more meaningful way. 105

106 Frequency Technical Mathematics Measurement of widgets produced (target measurement: 2.00 mm) Frequency Table Measurement interval Frequency In the frequency table above, the representation helps to form a clearer picture of what is happening. It can been seen that most of the widgets in this set are close to the target measurement and as the measurement moves away from the target, fewer and fewer of the widgets fall into that category. Another way to present data in a more graphical form is to use what is known as a histogram. Histograms are bar charts that represent the frequency distribution in a way that appeals to many readers. While histograms can be created by hand, there are many tools that can help to create wellformatted and visually appealing charts. The chart below was created in Microsoft Excel. Analysing widget production Measurement of widget in mm 106

107 Example 1 For the following data set, build the frequency table and the histogram using six 1-unit intervals. Solution: The data is tallied and categorized to form the frequency table: The data is now graphed to form the following histogram: 107

108 Frequency Technical Mathematics Analysing student experiment results Measurement of acceleration due to gravity 108

109 9.2 Measures of central tendency When analyzing data sets, there is often a need to have a single number that indicates a middle value. This can give some indication of the overall tendency of the data set. Common examples are the average score of a standardized test, the average salary of a given population, or the average measurement of a widget in a production process. These measurements are referred to as measurements of central tendency. There are three important such measurements as follows: Measure of central tendency Notation (if applicable) Description/formula mean x The sum of all of the data points divided by the number of data points x x x x x n n median Middle number of an ordered data set, or the mean of the two middle numbers if there are an even number of data points mode Most frequently occurring data point (there may be none or more than 1 mode) 109

110 Example 1 Find the mean, median, and mode of the following set of ACT scores: 17, 17, 17, 17, 18, 18, 19, 19, 20, 20, 21, 21, 22, 23, 25, 25, 26 Solution: Finding the mean: x 17 x Finding the median: Since the data set is an ordered data set with 17 data points, the 9 th element is the median Median = 20 Finding the mode: Since 17 occurs 4 times, and this is more than any other element, 17 is the mode Mode = 17 Analyzing these results it can be seen that all 3 of the measurements give a reasonable representation of the central tendency of this set. 110

111 Example 2 Find the mean, median, and mode of the following house prices: 70, ,000 72, ,000 81, ,000 90, ,000 94,000 8,500,000 Solution: x Finding the mean: 70, ,000+72, ,000+81, ,000+90, ,000+94,000+8,500, x 937, 600 Finding the median: Since the data set is an ordered data set with 10 data points, the mean of the 5 th and 6 th elements is the median Median = 94, , ,000 2 Finding the mode: Since all of the numbers occur exactly once (i.e. no number occurs more than any other number), this set does not have a mode Analyzing these results, the following can be seen: 1. Although the mean was calculated correctly, it is too high to be meaningful in this case. This is caused by the one house that had a value dramatically higher than all of the other houses. Data points like this are often called outliers and are sometimes left of calculations because of the skewing affect they have. 2. Since there is no mode, this calculation is not helpful at all. 3. The median is the only meaningful measure of central tendency for this set of data. 111

112 9.3 Measures of dispersion Whereas measures of central tendency attempt to provide the single number that indicates a middle value, sometimes it is important to know how dispersed or spread out the data set is. This value will often be used in many of the same circumstances as above to provide more insight into the data set. These measurements are referred to as measurements of dispersion. There are three important such measurements as follows: Measure of dispersion Notation (if applicable)* Description/formula range The largest number in the set minus the smallest number in the set variance s 2 n i1 ( x x ) n 1 i 2 standard deviation s n i1 ( x x ) n 1 i 2 * Sometimes notations other than s are used, but for simplicity this text will use s. 112

113 Example 1 Find the range, variance, and standard deviation of the set of ACT scores: 17, 17, 17, 17, 18, 18, 19, 19, 20, 20, 21, 21, 22, 23, 25, 25, 26 Solution: Finding the range: range Finding the variance: s n 2 i1 2 ( x xi ) ( ) ( )... ( ) n s Finding the standard deviation: s n i1 ( x x ) n 1 i 2 s ( ) ( )... ( ) s 9.10 s

114 The Windows OS calculator (as well as many other statistical calculators) has built in statistical functionality which can make these calculations much easier. The image below shows what the calculator looks like in statistical mode. The important keys for this discussion are the mean and standard deviation keys (circled in red). 114

115 Example 2 Use the calculator to find the standard deviation of the following house prices: 70, ,000 72, ,000 81, ,000 90, ,000 94,000 8,500,000 Solution: The data is added to the statistical calculator and the standard deviation is found. 115

116 9.4 Normally distributed data sets Some data sets will have their data naturally categorize into intervals according to a very standard system. The data sets are called normally distributed data sets, and occur frequently in both nature and artificial processes. Normally distributed data sets must meet the following structure (a small amount of error is allowed in artificial processes): 50% of the data is above the mean, and 50% is below The data is symmetric when graphed 68.2% of the data falls within one standard deviation of the mean 95.4% of the data falls within two standard deviations of the mean 99.7% of the data falls within three standard deviations of the mean The following graph encapsulates this structure: Standardized test scores are typically normally distributed, and analysis can be done based on this fact as in the following example. 116

117 Example 1 A standardized exam has the following characteristics: mean: 57 standard deviation: 9 If 250 students took this exam, how many scored between 48 and 66? Solution: 48 is one standard deviation below the mean (57 9 = 48) 66 is one standard deviation above the mean ( = 66) This question amounts to asking how many scores were within one standard deviation of the mean. Since this is a standardized exam (meaning it is normally distributed), it is known that 68.2% of the scores will be within this range. The following calculation generates the number: 250 (.682) = (rounded to 171) Hence, 171 students scored between 48 and

118 Example widgets are produced in a manufacturing process with a weight that follows a normal distribution according to the following characteristics: mean: 3.71 g standard deviation: g a) How many widgets weigh between g and g? b) How many widgets weigh over g? Solution: a) = (0.003) = = (0.003) = x 2s x 2s This question amounts to asking how many widgets weights are within two standard deviations of the mean. Since this is a normally distributed set, it is known that 95.4% of the weights will be within this range. The following calculation generates the number: 9000 (.954) = 8586 Hence, 8586 of the widgets weigh between g and g. b) = (0.003) = x 3s This question amounts to asking how many widgets weights are more than three standard deviations above the mean. Since this is a normally distributed set, it is known that 99.7% of the weights will be within three standard deviations, meaning only 0.3% will not. Of these 0.3%, half (or 0.15%) will be higher than x 3s. The following calculation generates the number: 9000 (.0015) = 13.5 (rounded up to 14) Hence, 14 of the widgets weigh more than g. 118

119 9.5 Controlled and capable processes In manufacturing, it is important to have a process work in such a way that all widgets are produced according to some specific control limits. While common cause variation prevents widgets from all being produced according to exact specifications, processes can be refined so that the widgets fall within certain parameters. When a process is refined to this point, it is called a controlled process. Controlled process when the measurements of all widgets produced fall within specified control limits (note this definition is a bit of an oversimplification but is precise enough for the discussion in this text). When an order is made by a customer, it will often come with tolerance limits (this represents the fact that the customer understands common cause variation and has identified the limits of measurement error they can tolerate). If a process is known to have control limits that fall within the tolerance limits, it is called a capable process. Capable process a controlled process where the control limits are within a customer s specified tolerance limits. Example 1 A controlled process produces widgets with the following characteristics (the target measurement is 7.5 cm): Upper control limit (UCL): cm Lower control limit (LCL): cm A customer order is entered with the following tolerance limit specification: 7.5 ± cm Is this process capable? Solution: The situation can be depicted by the following graphic: UTL = 7.503: UCL = 7.501: LCL = 7.499: LTL = 7.497: Since the control limits are within the tolerance limits (in other words, all widgets will be within the customer s tolerance), this is considered a capable process. The order can be accepted and filled. 119

120 Example 2 A controlled process produces widgets with the following characteristics (the target weight is 110 kg): Control limits: ± 5 kg A customer order is entered with the following tolerance limit specification: 110 ± 3.5 kg Is this process capable? Solution: The situation can be depicted by the following graphic: UCL = 115: UTL = 113.5: LTL = 106.5: LCL = 105: Since the control limits are not within the tolerance limits, it cannot be guaranteed that all widgets will be within the customer s tolerance. Hence, this is not a capable process. The order cannot be accepted. 120

121 Chapter 9 Homework problems Section 9.1 Homework 1. In a manufacturing process, a new starter is being installed into a product during assembly. Quality control personnel are tracking how many of the starters are failing each day over the course of a month. For the following data set, build the frequency table and the histogram using four 5-unit intervals. Start the lowest interval with 2 (so the first interval will be 2 6). 2. The following test scores were collected over a 5-year period. The grading scale was a straight grading scale to indicate A, B, C, and D grades respectively (anything below a 60 is considered a failing grade). Create the frequency table and histogram using 4 intervals one for each of the passing grades. 121

122 Section 9.2 Homework For questions 1 3, find the mean, median, and mode. After finding those measures, discuss whether any of them are a better reflection of the central tendency for the data set. 1. 4, 7, 9, 9, 9, 14, 14, 16, 23, 31, , , 1, 1, 2, 2, 2, 3, 3, 3, 3, 3, 5, 5, 5, 6 4. Is the order of the data set important when calculating the mean? The median? The mode? Section 9.3 Homework 1. Manually calculate the range and the standard deviation of the following data set: 3, 5, 7, 7, 9 2. Use a statistical calculator to verify the standard deviation calculated in question Use a statistical calculator to calculate the range and standard deviation of the following data sets: a. 17, 18, 18, 22, 44, 45, 45, 45, 65, 71, 73, 78, 80, 80, 91, 92, 93, 99 b. 100, 300, 250, 175, 188, 337, 299, 171, 152, Is the order of the data set important when calculating the range? The standard deviation? 122

123 Section 9.4 Homework ,000 widgets are manufactured with a target measurement of 11.5 cm. After the widgets are measured and the data is collected, the distribution characteristics are: x cm s cm a. How many of the widgets measure between cm and 11.5 cm? b. How many of the widgets measure between cm and cm? c. How many of the widgets measure between and cm? d. How many of the widgets measure between and cm? e. How many of the widgets measure at least cm? f. If an order is placed for 15,000 widgets, but only under the condition that at least 90% of them are between cm and cm, can this order be accepted? 2. Professional baseball batting averages that fell between.280 and.289 were tracked over a season. The distribution of a sampling of 10 players is: 0.280, 0.282, 0.282, 0.284, 0.285, 0.287, 0.287, 0.287, 0.289, Assuming this data is normally distributed, use the sampling to determine the pertinent characteristics of the data set, and then determine how many of the 73 players who had a percentage between.280 and.289 were either above or below

124 Section 9.5 Homework 1. A controlled process produces widgets with the following characteristics (the target measurement is 33.2 mm): Upper control limit (UCL): cm Lower control limit (LCL): cm A customer order is entered with the following tolerance limit specification: 33.2 ± 0.01 cm. Is this process capable? 2. Another customer order for the widgets in question 1 is entered with the following tolerance limit specification: 33.2 ± cm. Is this process capable? 3. A controlled process produces widgets with the following characteristics (the target weight is 100 kg): Control limits: ±2.5 kg A customer order is entered with the following tolerance limit specification: ± 3.3 (units are measured in lb) Is this process capable? 124

125 Chapter 10 Basic Logic 10.1 Logical expressions Logical expressions revolve around situations that have only one of two ways to be resolved. These situations are often referred to as Boolean scenarios, and are not entirely unrelated to the binary discussion in the Number systems chapter. Situations in technical settings often revolve around circuitry or computer programming. They also occur in many other areas. A small sample of situations in technology are given below. In circuitry the Boolean values occur in situations such as: o o Determining if the voltage is high or low Determining whether a given scenario occurs, and taking action based on that scenario (for example, if a car is waiting in a left turn lane the logic controller will detect that and give a green turn arrow when the time comes) In computer programming the Boolean values occur and dictate the logic flow: o o The value may determine what module is executed (for example, if the employee is a manager one routine might be executed, otherwise some other routine might be executed) The value may determine how many times a module is executed (for example, as long as there is data in a file the records are processed and as soon as there is no more data the processing ends) Evaluating logical expressions is very similar to evaluating algebraic expressions. Recall that in algebraic expressions, we use the operations +,, x, and. Example 1 (Reviewing algebraic expressions) If x = 3, y = 2, and z = 11, evaluate the algebraic expression: xy y z. Solution: 2 2xy 3y 4 z when x 3, y 2, and z (3)(2) 3(2) 4(11)

126 A similar process is used to evaluate logical expressions. However, the operations are no longer the four familiar arithmetic operations. The logical operators are: Boolean operator Notation Description And Evaluates to true if both operands are true Or Evaluates to true if either operand (or both) is true Not (the single apostrophe) Inverts the value of the operand These three operators are more formally defined by their truth tables. A truth table provides all of the necessary information for a logical operator. The And logical operator When logical variables are needed, p, q, and r are often used (as opposed to x, y, and z in algebra). Additional, the two values used are typically 0 and 1, and these stand for the two states in the system (the states might be 1 = high voltage and 0 = low voltage in an electronics context, 1 = true and 0 = false in a programming context, etc.) The And logical operator is defined by the following truth table (where p and q are Boolean variables): p q p ^ q This allows for expressions to be evaluated. Example 2 If p = 1 and q = 0, evaluate the logical expression: ( p q) p. Solution: In a similar fashion to algebraic expressions, logical expression are evaluated by substituting in the appropriate data, performing the operations, and obtaining the final value. ( p q) p when p 1 and q 0 (1 0)

127 The Or logical operator The Or logical operator is defined by the following truth table: p q p q Example 3 If p = 1 and q = 0, evaluate the logical expression: ( p q) p. Solution: In a similar fashion to algebraic expressions, logical expression are evaluated by substituting in the appropriate data, performing the operations, and obtaining the final value. ( p q) p when p 1 and q 0 (1 0) The Not logical operator The Not logical operator is defined by the following truth table (since the Not operator simply inverts the operand, only one variable is needed): p p

128 10.2 Logic truth tables Now that the three fundamental operations have been defined, larger truth tables that encompass more complicated situations can be formed. Again, there are many scenarios in technology where this process is needed. A couple of examples would be: In the stop light example above, the controller will have to sense many possible situations. For example, if there is a car in one left turn lane and a car in the opposite turn lane, then provide both of them with left turn arrow, if there is only a car in one turn lane then provide the cars on that side of the street with both a green arrow and a green light, if there is not a car in either turn lane, then just provide a green light, etc. In the programming example above, it may be a case where one module is executed if the employee is a manager and has a salary above a certain value or if they are an administrator, another module might be executed if they are a programmer, analyst, or DBA, and yet a third module is executed in all other cases. In these situations, the logic device will have to evaluate the particular expression and execute the appropriate commands. Truth tables allow designers to lay out all of the possibilities so all scenarios can be expressed in one structure. This allows for analysis and problem-solving of the given situation. Truth tables for logical expressions will be tables with a certain number of rows and columns. They are created by following this algorithm (an algorithm is a step-by-step step set of instructions for accomplishing some task). 1. Create columns for each of the variables in the expression, in alphabetical order 2. Use additional columns to build up to the expression, one operation at a time 3. Create 2 n rows, where n is the number of unique variables in the expression 4. Enter all possible combinations of initial conditions in the columns for the variables 5. For all of the remaining columns, fill them in one at the time applying the appropriate rules for the current operation The following example shows every step in this process: Example 4 Create a truth table for the logical expression: ( p q)' p'. Solution: Steps 1, 2, and 3 form the structure of the truth table p q (p q) (p q) p (p q) p 128

129 Step 4 provides all combinations of initial conditions (these are the first 2 columns) p q (p q) (p q) p (p q) p Step 5 is done one column at a time, focusing on the current operands and operator. Column 3: p q (p q) (p q) p (p q) p Column 4: p q (p q) (p q) p (p q) p Column 5: p q (p q) (p q) p (p q) p And finally column 6 to complete the table: p q (p q) (p q) p (p q) p 129

130 Analyzing the results: While all of the columns are an important part of the process, the most important column is the last one. This provides the ultimate result for this expression given a set of initial conditions. For example if p = 0 and q = 1, then the expression evaluates to 0. This can be seen by identifying row 2 as the row associated with the initial conditions p = 0 and q = 1, and then looking to the far right column to see the result of 0. Typically the table is not replicated for each new column. The steps above are followed, but only one table is needed as demonstrated in the next example. Example 5 Create a truth table for the logical expression: ( p' r) ( q p'). Solution: p q r p (p r) (q p ) (p r) (q p ) Analyzing the results: Since there are 3 variables, 2 3 = 8 rows are needed to handle all possible combinations of initial conditions. Other than that, the table works in the same way as any other table. 130

131 10.3 Schematics Logical expressions can be expressed as in sections 10.1 and 10.2, or then can be expressed using technical schematics. When schematics are used, the three fundamental operations are defined and notated as in the following table: Operator Image Image name And And gate Two inputs, one output OR Or gate Two inputs, one output Not Inverter One input, one output Building the actual schematic follows a similar process as choosing the columns in the corresponding truth table. The following 2-step algorithm indicates what must be done: 1. Write each of the variables in the expression vertically along the left-hand side of the schematic 2. Build the schematic one operation at a time, thinking through and indicating the inputs to each gate/inverter 131

132 Example 1 Create the logic schematic for the logical expression from the example above: ( p q)' p'. Solution: Following the two steps above, the following schematic is created: p q Schematics can also be used to generate a logical expression. It is important to read the schematics from left to right to obtain the correct expression. Example 2 Determine the logical expression associated with the following schematic: Solution: Reading the schematic from left to right, the following expression is determined: ( p q)' (( q r) r) 132

133 Chapter 10 Homework problems Section 10.1 Homework 1. If x, y, and z all have the value 1, evaluate the logical expression: x ( y z '). 2. If x is 1 and y is 0, evaluate the expression: ( x y ') ( y ( x' y)) Section 10.2 Homework 3. Construct the truth table for the expression in question Construct the truth table for the expression in question 2. Section 10.3 Homework 5. Construct the logic schematic for the expression in question Construct the logic schematic for the expression in question 2. Section 10.4 Homework 7. Determine the logical expression associated with the following schematic: p q 133

134 8. Determine the logical expression associated with the following schematic: 134

135 Supplemental Chapter Systems of Linear Equations S.1 Solving systems of 2 linear equations using a graphing tool There are often times in technological settings when two linear equations (which would graph as lines) are part of a system. When this happens, certain applications require knowing when the two lines intersect, as this point of intersection would provide the (often unique) solution to the system. For example, consider the following two lines: y3x2 y 1 2 x4 When graphed on the TI-83, these two lines look as follows: If the point of intersection needs to be determined, it can be done on the calculator using the intersect functionality resulting in the following image: The intersection point calculated by the TI-83 is (2.4, 5.2). What this means is that while there are an infinite number of points that satisfy the line y3x 2, and an infinite number of points that satisfy 1 the line y x, there is ONE AND ONLY ONE point that satisfies both of them specifically the point 2 4 (2.4, 5.2) referenced above. 135

136 The following example illustrates the use of a TI-83 to solve a system in more detail. Example 1 Solve the following system of equations using a TI-83 y 1.5x1 y2x3.2 Solution: 1. Enter the equations into the calculator (press the Y= button to do this) 2. Graph the equations (press the GRAPH button to do this) 3. Use the TI-83 to calculate the point of intersection by doing the following: Press 2nd' and then TRACE Press 5 to choose intersect Press ENTER to choose the first graph Press ENTER to choose the second graph Move the cursor close to the point of intersection and press ENTER 136