28: Topic D: Formulas—Computing and Graphing - Mathematics

28: Topic D: Formulas—Computing and Graphing - Mathematics

Math Formulas for Geometric Shapes

In math (especially geometry) and science, you will often need to calculate the surface area, volume, or perimeter of a variety of shapes. Whether it's a sphere or a circle, a rectangle or a cube, a pyramid or a triangle, each shape has specific formulas that you must follow to get the correct measurements.

We're going to examine the formulas you will need to figure out the surface area and volume of three-dimensional shapes as well as the area and perimeter of two-dimensional shapes. You can study this lesson to learn each formula, then keep it around for a quick reference next time you need it. The good news is that each formula uses many of the same basic measurements, so learning each new one gets a little easier.

GED Math Practice Test with Answers 2020

Q1. The dimensions of Box B, shown below, are twice the length of the corresponding dimensions on Box A (not shown).

  • A). The volume of Box A is greater than the volume of Box B.
  • B). The volume of Box B is twice the volume of Box A.
  • C). The volume of Box B is three times the volume of Box A.
  • D). The volume of Box B is four times the volume of Box A.
  • E). The volume of Box B is eight times the volume of Box A.

Since the dimensions of Box A are half of the dimensions of Box B, the side lengths must be 3, 2, and 1.5. Next, find the volumes of the two
boxes. Use the formula V = lwh. The volume of Box B is 72, and the volume of Box A is 9. 72 is 8 times larger than 9.

Q2. At the end of baseball season, 5% of the children enrolled in a local youth baseball program will be chosen to play in the state tournament. If 12 children will be chosen to play in the tournament, how many children are enrolled in the

Set up a proportion: 5/100 = 12/x , where x is the number of children enrolled in the program. 12 × 100 = 1,200, and 1,200 ÷ 5 = 240.

Q3. A bag contains 12 red, 3 blue, 6 green, and 4 yellow marbles. If a marble is drawn from the bag at random, what is the probability that the marble will be either blue or yellow?

Add the number of marbles to get the total number in the bag 12 + 3 + 6 + 4 = 25. Therefore, 25 is the number of possible outcomes.
Seven marbles are either blue or yellow. Seven is the number of favorable outcomes (7/25) x (4/4) =28/100 =28%

Q4. Patricia wants to order business cards. A printing company determines the cost (C) to the customer using the following function, where b = the number of boxes of cards and n = number of ink colors.
C = $25.60b + $14.00b(n − 1) If Patricia orders 4 boxes of cards printed in 3 colors, how much will the cards cost?

Substitute 4 for b and 3 for n into the function. Then, solve the equation.
C = $25.60(4) + $14(4)(3 − 1)
= $102.40 + $112.00
= $214.40

Q5. Andrea bought a used mountain bike for $250. She gave the bike a new paint job replaced the tires, chain, and gear assembly and sold the bike for 150% of the price she paid. For what amount, in dollars, did she sell the bike?

Multiply $250 by 1.5, which equals $375.

Q6. Ajani finds that the distance between two landmarks on a map is 6 (½) inches. If the map scale reads ¾ inch = 120 miles, what is the actual distance, in miles, between the two landmarks?

Set up the proportion and solve: x = the number of miles between the two landmarks.

Q7. Which of the following is a graph of the inequality −2 ≤ x < 4?

Separate the two inequalities into two inequalities, x < 4 and −2 = x. Choice d is the only graph that represents the inequalities. There must be an open circle to represent that the 4 is not included and a shaded circle to represent that the −2 is included.

Questions 8 and 9 are based on the following figure.

Lines A and B are parallel.

Q8. What is the measure of ∠4?

Angle 4 and the angle measuring 50° are corresponding angles. Therefore,m∠4 = 50°.

Q9. What is the value of x?

Angle 3 and the angle measuring 104° are supplementary therefore,m∠3 = 180 − 104 = 76. As established in the previous problem, angle 4
and the angle measuring 50° are corresponding angles, so m∠4 = 50°. Angle 3, angle 4, and x are the interior angles of a triangle, so they equal 180° 50 + 76 + x = 180, and so x = 54°.

Q10. On a coordinate plane, a vertical line is drawn through the point (−3,4). On the same plane, a horizontal line is drawn through the point (2,−1). At what point on the plane will the two lines intersect?

The vertical line is parallel to the y-axis, and all of its points have the x-coordinate −3. The horizontal line is parallel to the x-axis, and all of its points have the y-coordinate −1. Therefore, the coordinates are −3 and −1.

Q11. What was Edmundo’s mean score for a round of golf in August if his scores for each round were 78, 86, 82, 81, 82, and 77?

Mean = average. Add the scores and divide by the number of scores. 78 + 86 + 82 + 81 + 82 + 77 = 486 => 486 ÷ 6 = 81

Q12. In quadrilateral ABCD, side AB is parallel to side CD. Sides AD and BC are not parallel. What is the area of the figure to the nearest square centimeter?

Quadrilateral ABCD is a trapezoid because it has one pair of parallel sides. The bases are the parallel sides, AB and CD. The height is 2.5 cm.
Use the formula for the area of a trapezoid.

Q13. A display of canned yams has a sign that reads “5 cans for $3.” At the same rate, how much would the store charge, to the nearest whole cent, for 8 cans?

Set up the proportion 5/3 = 8/x , where x is the cost of 8 cans of yams 3 × 8 = 24, and 24 ÷ 5 = $4.80.

Q14. A plastic pipe, 5 feet 9 inches long, is cut into three equal pieces. Assuming no waste when the cuts are made, what is the length of each piece?

5 ft. 9 in. = 69 inches. Divide by 3: 69 ÷ 3 = 23 inches. Convert feet to inches, 23 in. = 1 ft. 11 in.

Q15. The Wrights plan to carpet their bedroom and the adjoining hallway as shown in the diagram. What is the area of the bedroom and hallway in square feet?

Think of the space as two rectangles: 21 ft. by 14 ft. and 10 ft. by 7 ft. (You can find the length of the missing side of the smaller rectangle by subtracting: 24 − 14 = 10.) Use the formula A = lw to find the area of each rectangle, and then combine to find the total area of the floor to be carpeted: 21 × 14 = 294, and 7 × 10 = 70. Add the areas of the two rectangles: 294 + 70 = 364, so the Wrights will need to buy 364 square feet of carpeting.

Q16. One number is 12 more than 3 times another number. The sum of the 2 numbers is −20. What are the numbers?

x = one number and 3x + 12 = the other number,
for the equation:
x + 3x + 12 = − 20
4x + 12 = −20
4x = −32
x = −8
3(−8) + 12 = −12

Questions 17 and 18 refer to the following graph.

The Kleins are trying to pay off their credit card debt, so they developed the following budget based on their monthly take-home pay.

Q17. If the Kleins’ monthly take-home pay is $2,500, about how much do they plan to pay each month on their credit card debt?

Q18. What fraction of the Kleins’ monthly take-home pay goes toward clothing?

Clothing expenses take 5% of the Wrights’ pay. Change 5% to a fraction to get 5/100, which can also be reduced to 1/20.

Q19. The graph of the equation y = − (3/4) x + 1 is a line that passes through points C and D on the coordinate plane. Which of the following points also lies on the graph of the equation?

Either locate each point on the grid and compare it to the line or substitute the x and y values from each ordered pair into the equation:

y = − ¾ x + 1
−5 = − ¾ (8) + 1
−5 = −6 + 1
−5 = −5

Q20. The sum of three consecutive even integers is 90. What is the greatest number in the series?

The three numbers can be represented by x,
x + 2, and x + 4. Solve the equation:
x + x + 2 + x + 4 = 90
3x + 6 = 90
3x = 84
x = 28

The three numbers are 28, 30, 32. The question asks for the largest of these.

Q21. The three interior angles of ΔJKL measure 45°, 45°, and 90°. The three interior angles of ΔPQR measure 45°, 45°, and 90°. The side JK of JKL measures 4 cm. Based on the given information, which of the following must be a true statement?

  • A). Δ JKL and ΔPQR are acute triangles.
  • B). One side of ΔPQR measures 4 cm
  • C). The perimeter of ΔJKL is 12 cm.
  • D). ΔJKL and ΔPQR are congruent triangles.
  • E). ΔJKL and ΔPQR are similar triangles.

Similar triangles have the same angle measures. Congruent triangles have the same angle measures and the same side lengths. From the given information, you cannot know if the side lengths are the same, so you can conclude only that the triangles are similar.

Q22. A parallelogram is drawn on a coordinate grid so that three vertices are located at (3,4), (−2,4), and (−4,1). At what coordinates should the fourth vertex be located?

Plot the points given in the problem and compete with the parallelogram. Remember that in a parallelogram, both pairs of opposite sides are
equal and parallel.

Q23. Eight athletes ran a 1 mile race. The winner’s time was 4 minutes 8 seconds. If the median time was 4 minutes 48 seconds, what was the time of the athlete who finished last?

  • A). 5 min. 28 sec.
  • B). 5 min. 4 sec.
  • C). 4 min. 46 sec.
  • D). 4 min. 28 sec.
  • E). Not enough information is given.

The median time is the middle time. There is no way of knowing how far behind the median the slowest runner was.

Questions 24 and 25 refer to the following information.

Q24. On what date and in what location was there the greatest jump in the price of gasoline from one week to the next?

  • A). April 23 on the West Coast
  • B). April 30 in the Midwest
  • C). April 30 on the West Coast
  • D). May 7 on the East Coast
  • E). May 7 in the Midwest

The steepest rise on the graph was from April 23 to April 30. The symbol indicates that it was in the Midwest.

Q25. Based on the information in the graph, which of the following is the best prediction of the price per gallon of gasoline on the West Coast for the week following May 7?

The prices for the West Coast have been rising steadily by 2 or 3 cents each week. On May 7, the price on the West Coast is a little beneath $1.80. If it rises 2 or 3 cents, it should be at about $1.82 by the following week. The question gives no reason to expect a sudden decline in price or a sharp increase.

Q26. Chikita made three long-distance calls. According to her phone bill, the calls were 19 minutes, 24 minutes, and 8 minutes in length. If Chikita pays 9 cents per minute on all long-distance calls, how much was she billed for the three calls?

Add the times and multiply by 9 cents: 19 + 24 + 8 = 51 minutes 51 × .09 = $4.59.

Q27. Maggie and Christian decided to share the cost of buying their friends a wedding gift. Maggie put in $20 less than twice the amount that Christian contributed. Together, they spent $94. How many dollars did Maggie contribute toward the gift?

Let x represents the amount that Christian put in and 2x − 20 represents Maggie’s contribution. Solve the equation.
x + 2x − 20 = 94
3x = 114
x = 38
Christian put in $38 and Maggie put in 94 − 38 = $56.

Q28. The Northridge Quakers have won 20 games and lost 15. What is the ratio of games won to game played?

The number of games played is the total of the wins and losses, (20 + 15 = 35). Write the ratio and simplify. 20/35 = 4/7

Q29. To set up a tent, workers place a 25-foot pole in the center of a grassy area as shown in the diagram. A bracing wire is attached to the top of the pole and to a stake 9 feet from the base of the pole. Which of the following represents the
length of the bracing wire?

Q30. What is the value of the expression −3 × 5² + 2(4 − 18) + 3²?

−3 × 5² + 2(4 − 18) + 3³
−3 × 25 + 2(−14) + 27
−75 + (−28) + 27

Q31. The lengths of the sides of ΔABC are 6 inches, 8 inches, and 10 inches. Which of the following conclusions must be true?

  • A). ∠C is a right angle
  • B). ΔABC is an acute triangle.
  • C). ΔABC contains one obtuse angle.
  • D). ∠A is an acute angle.
  • E). m∠A + m∠B + m∠C = 180°

ABC is a right triangle, but you have no way of knowing which angle is a right angle, so eliminate choice a. Regardless of the type of triangle,
the sum of the measure of the interior angles of the triangle must be 180°.

Q32. Two sides of a triangle measure 6 and 10 inches. If the triangle is a right triangle, which of the following could be the measure, in inches, of the third side?

You can try each combination of sides in the Pythagorean theorem. Only 6, 8, and 10 will work. 62 + 82 = 102, 36 + 64 = 100, 100 = 100.

Q33. Classify the figure containing interior angles 3, 6, 11, and 13.

  • A). scalene triangle
  • B). trapezoid
  • C). parallelogram
  • D). rectangle
  • E). square

A four-sided figure with only one pair of parallel sides is a trapezoid.

Q34. The floor of a walk-in closet measures 7 feet by 4 feet. If the ceiling height is 8 feet, what is the volume in cubic feet of the closet?

A closet is the shape of a rectangular solid. To find the volume, multiply. V = lwh. 7(4)(8) = 224 cubic feet.

Q35. In a right triangle, the hypotenuse measures 15 inches. If one leg of the triangle measures 6 inches, which of the following equations could be used to find the length of the other leg (x) in inches?

Use the Pythagorean theorem. If a² + b² = c² and c is the hypotenuse, then b² = c² − a² or x² = 15² − 6².

Table of Contents

Mini-Mod 1 Whole Numbers
Topic 1.1 Whole Numbers
Topic 1.2 Rounding
Topic 1.3 Adding Whole Numbers Estimation
Topic 1.4 Subtracting Whole Numbers
Topic 1.5 Basic Problem Solving
Topic 1.6 Multiplying Whole Numbers
Topic 1.7 Dividing Whole Numbers
Topic 1.8 More with Multiplying and Dividing
Topic 1.9 Exponents
Topic 1.10 Order of Operations and Whole Numbers
Topic 1.11 More Problem Solving

Mini-Mod 2 Factors and Fractions
Topic 2.1 Factors
Topic 2.2 Prime Factorization
Topic 2.3 Understanding Fractions
Topic 2.4 Simplifying Fractions - GCF and Factors Method
Topic 2.5 Simplifying Fractions - Prime Factors Method
Topic 2.6 Multiplying Fractions
Topic 2.7 Dividing Fractions

Mini-Mod 3 LCM and Fractions
Topic 3.1 Finding the LCM - List Method
Topic 3.2 Finding the LCM - GCF Method
Topic 3.3 Finding the LCM - Prime Factor Method
Topic 3.4 Writing Fractions with an LCD
Topic 3.5 Adding and Subtracting Like Fractions
Topic 3.6 Adding and Subtracting Unlike Fractions
Topic 3.7 Order of Operations and Fractions

Mini-Mod 4 Mixed Numbers
Topic 4.1 Changing a Mixed Number to an Improper Fraction
Topic 4.2 Changing an Improper Fraction to a Mixed Number
Topic 4.3 Multiplying Mixed Numbers
Topic 4.4 Dividing Mixed Numbers
Topic 4.5 Adding Mixed Numbers
Topic 4.6 Subtracting Mixed Numbers
Topic 4.7 Adding and Subtracting Mixed Numbers - Improper Fractions

Mini-Mod 5 Operations with Decimals
Topic 5.1 Decimal Notation
Topic 5.2 Comparing Decimals
Topic 5.3 Rounding Decimals
Topic 5.4 Adding and Subtracting Decimals
Topic 5.5 Multiplying Decimals
Topic 5.6 Dividing Decimals
Topic 5.7 Order of Operations and Decimals
Topic 5.8 Converting Fractions to Decimals
Topic 5.9 Converting Decimals to Fractions

Mini-Mod 6 Ratios, Rates, and Percents
Topic 6.1 Ratios
Topic 6.2 Rates
Topic 6.3 Proportions
Topic 6.4 Percent Notation
Topic 6.5 Percent and Decimal Conversions
Topic 6.6 Percent and Fraction Conversions
Topic 6.7 The Percent Equation
Topic 6.8 The Percent Proportion
Topic 6.9 Percent Applications

Mini-Mod 7 U.S. and Metric Measurement
Topic 7.1 U.S. Length
Topic 7.2 U.S. Weight and Capacity
Topic 7.3 Metric Length
Topic 7.4 Metric Mass and Capacity
Topic 7.5 Converting between U.S. and Metric Units
Topic 7.6 Time and Temperature

Mini-Mod 8 Introduction to Geometry
Topic 8.1 Lines and Angles
Topic 8.2 Figures
Topic 8.3 Perimeter - Definitions and Units
Topic 8.4 Finding Perimeter
Topic 8.5 Area - Definitions and Units
Topic 8.6 Finding Area
Topic 8.7 Understanding Circles
Topic 8.8 Finding Circumference
Topic 8.9 Finding Area - Circles

Mini-Mod 9 More Geometry
Topic 9.1 Volume - Definitions and Units
Topic 9.2 Finding Volume
Topic 9.3 Square Roots
Topic 9.4 The Pythagorean Theorem
Topic 9.5 Similar Figures
Topic 9.6 Finding Missing Lengths

Mini-Mod 10 Statistics
Topic 10.1 Bar Graphs
Topic 10.2 Line Graphs
Topic 10.3 Circle Graphs
Topic 10.4 Mean
Topic 10.5 Median
Topic 10.6 Mode

Mini-Mod 11 Real Numbers and Variables
Topic 11.1 Introduction to Real Numbers
Topic 11.2 Graphing Rational Numbers Using a Number Line
Topic 11.3 Translating Phrases into Algebraic Inequalities
Topic 11.4 Finding the Absolute Value of a Real Number

Mini -Mod 12 Adding and Subtracting with Real Numbers
Topic 12.1 Adding Real Numbers with the Same Sign
Topic 12.2 Adding Real Numbers with Different Signs
Topic 12.3 Finding the Opposite of a Real Number
Topic 12.4 Subtracting Real Numbers
Topic 12.5 Addition Properties of Real Numbers

Mini-Mod 13 Multiplying and Dividing with Real Numbers
Topic 13.1 Multiplying Real Numbers
Topic 13.2 Finding the Reciprocal of a Real Number
Topic 13.3 Dividing Real Numbers
Topic 13.4 Exponents and the Order of Operations
Topic 13.5 The Distributive Property
Topic 13.6 Multiplication Properties of Real Numbers

Mini-Mod 14 Variables and Expressions
Topic 14.1 Introduction to Expressions
Topic 14.2 Evaluating Algebraic Expressions
Topic 14.3 Simplifying Expressions
Topic 14.4 Simplifying Expressions with Parentheses
Topic 14.5 Translating Words into Symbols

Mini-Mod 15 Introduction to Solving Linear Equations
Topic 15.1 Translating Words into Equations
Topic 15.2 Linear Equations and Solutions
Topic 15.3 Using the Addition Property of Equality
Topic 15.4 Using the Multiplication Property of Equality
Topic 15.5 Using the Addition and Multiplication Properties Together

Mini-Mod 16 Solving More Linear Equations and Inequalities
Topic 16.1 Solving Equations with Variables on Both Sides
Topic 16.2 Solving Equations with Parentheses
Topic 16.3 Solving Equations with Fractions
Topic 16.4 Solving a Variety of Equations
Topic 16.5 Solving Equations and Formulas for a Variable
Topic 16.6 Solving and Graphing Linear Inequalities

Mini- Mod 17 Introduction to Graphing Linear Equations
Topic 17.1 The Rectangular Coordinate System
Topic 17.2 Graphing Linear Equations by Plotting Points
Topic 17.3 Graphing Linear Equations Using Intercepts
Topic 17.4 Graphing Linear Equations of the Form x=a, y=b, and y=mx

Mini - Mod 18 Slope and the Equation of a Line
Topic 18.1 The Slope of a Line
Topic 18.2 Slope-Intercept Form
Topic 18.3 Graphing Lines Using the Slope and y-Intercept
Topic 18.4 Writing Equations of Lines Using a Point and Slope
Topic 18.5 Writing Equations of Lines Using Two Points

Mini-Mod 19 Introduction to Functions
Topic 19.1 Relations and Functions
Topic 19.2 The Vertical Line Test
Topic 19.3 Function Notation
Topic 19.4 Evaluating Functions
Topic 19.5 Piecewise Functions

Mini-Mod 20 Solving Systems of Linear Equations
Topic 20.1 Introduction to Systems of Linear Equations
Topic 20.2 Solving by the Graphing Method
Topic 20.3 Solving by the Substitution Method
Topic 20.4 Solving by the Elimination Method

Mini-Mod 21 Introduction to Polynomials and Exponent Rules
Topic 21.1 Introduction to Polynomials
Topic 21.2 Addition of Polynomials
Topic 21.3 Subtraction of Polynomials
Topic 21.4 Product Rule for Exponents
Topic 21.5 Power Rule for Exponents

Mini - Mod 22 Multiplying Polynomials
Topic 22.1 Multiplying by a Monomial
Topic 22.2 Multiplying Binomials
Topic 22.3 Multiplying Polynomials
Topic 22.4 Multiplying the Sum and Difference of Two Terms
Topic 22.5 Squaring Binomials

Mini-Mod 23 Dividing Polynomials and More Exponent Rules
Topic 23.1 The Quotient Rule
Topic 23.2 Integer Exponents
Topic 23.3 Scientific Notation
Topic 23.4 Dividing a Polynomial by a Monomial
Topic 23.5 Dividing a Polynomial by a Binomial

Mini-Mod 24 Factoring Polynomials
Topic 24.1 Greatest Common Factor
Topic 24.2 Factoring by Grouping
Topic 24.3 Factoring Trinomials of the Form
Topic 24.4 Factoring Trinomials of the Form
Topic 24.5 More Factoring of Trinomials

Mini-Mod 25 More Factoring and Quadratic Equations
Topic 25.1 Special Cases of Factoring
Topic 25.2 Factoring Polynomials
Topic 25.3 Solving Quadratic Equations by Factoring
Topic 25.4 Applications

Mini-Mod 26 Introduction to Rational Expressions
Topic 26.1 Undefined Rational Expressions
Topic 26.2 Simplifying Rational Expressions
Topic 26.3 Multiplying Rational Expressions
Topic 26.4 Dividing Rational Expressions

Mini-Mod 27 Adding and Subtracting Rational Expressions
Topic 27.1 Adding Like Rational Expressions
Topic 27.2 Subtracting Like Rational Expressions
Topic 27.3 Finding the Least Common Denominator for Rational Expressions
Topic 27.4 Adding and Subtracting Unlike Rational Expressions

Mini-Mod 28 Complex Rational Expressions and Rational Equations
Topic 28.1 Simplifying Complex Rational Expressions by Adding and Subtracting
Topic 28.2 Simplifying Complex Rational Expressions by Multiplying by the LCD
Topic 28.3 Solving Rational Equations
Topic 28.4 Direct Variation
Topic 28.5 Inverse Variation
Topic 28.6 Applications

Mini-Mod 29 Roots and Radicals
Topic 29.1 Square Roots
Topic 29.2 Higher-Order Roots
Topic 29.3 Simplifying Radical Expressions
Topic 29.4 Rational Exponents
Topic 29.5 The Pythagorean Theorem
Topic 29.6 The Distance Formula

Mini-Mod 30 Operations of Radical Expressions
Topic 30.1 Adding and Subtracting Radical Expressions
Topic 30.2 Multiplying Radical Expressions
Topic 30.3 Dividing Radical Expressions
Topic 30.4 Rationalizing the Denominator
Topic 30.5 Solving Radical Equations

Mini-Mod 31 Solving Quadratic Equations
Topic 31.1 Introduction to Solving Quadratic Equations
Topic 31.2 Solving Quadratic Equations by Factoring
Topic 31.3 Solving Quadratic Equations using the Square Root Property
Topic 31.4 Solving Quadratic Equations by Completing the Square
Topic 31.5 Solving Quadratic Equations using the Quadratic Formula

Mini-Mod 32 Graphing Quadratic Equations
Topic 32.1 Introduction to Graphing Quadratic Equations
Topic 32.2 Finding the Vertex of a Quadratic Equation
Topic 32.3 Finding the Intercepts of a Quadratic Equation
Topic 32.4 Graphing Quadratic Equations Summary

Mini-Mod 33 Compound and Quadratic Inequalities
Topic 33.1 Sets
Topic 33.2 Interval Notation
Topic 33.3 Graphing Compound Inequalities
Topic 33.4 Solving Compound Inequalities
Topic 33.5 Solving Quadratic Inequalities

Mini-Mod 34 Absolute Value Equations and Inequalities
Topic 34.1 Introduction to Absolute Value Equations
Topic 34.2 Solving Basic Absolute Value Equations
Topic 34.3 Solving Multiple Absolute Value Equations
Topic 34.4 Solving Absolute Value Inequalities

Appendix A Conic Sections
Topic A.1 The Circle
Topic A.2 The Parabola
Topic A.3 The Ellipse
Topic A.4 The Hyperbola

Appendix B Logarithmic and Exponential Functions
Topic B.1 Evaluating Exponential and Logarithmic Expressions
Topic B.2 Graphing Exponential Functions
Topic B.3 Solving Simple Exponential Equations and Applications
Topic B.4 Converting Between Exponential and Logarithmic Forms
Topic B.5 Solving Simple Logarithmic Equations
Topic B.6 Graphing Logarithmic Functions

Appendix C Solving Logarithmic and Exponential Equations
Topic C.1 Properties of Logarithms
Topic C.2 Common and Natural Logarithms
Topic C.3 Change of Base of Logarithms
Topic C.4 Solving Exponential Equations and Applications
Topic C.5 Solving Logarithmic Equations and Applications

Appendix D Additional Topics
Topic D.1 The Midpoint Formula
Topic D.2 Writing Equations of Parallel and Perpendicular Lines
Topic D.3 Graphing Linear Inequalities in Two Variables
Topic D.4 Systems of Linear Inequalities
Topic D.5 Synthetic Division
Topic D.6 Complex Numbers
Topic D.7 Matrices and Determinants
Topic D.8 Cramer&rsquos Rule
Topic D.9 Solving Systems of Linear Equations Using Matrices

Topic D.10 Basic Probability and Statistics

Topic D.11 Applications of Systems of Linear Equations


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Mathematics and Sports

Some of the fascinating mathematics of sports scheduling.


Sports in America is both big business and democracy in action. Baseball, football, and basketball have become writ large, with games on television and cable television generating huge sums of money that goes into the American economy. But sports are also a mechanism whereby American children learn the values of fair play, hard effort, and the meaning of friendship.

Mathematics Awareness Month was created to help focus the public's attention on the nature of mathematics and the way it affects our daily lives and fosters the development of new tools to solve problems for individuals, businesses, and governments. This year the theme for Mathematics Awareness Month is Mathematics and Sports.

The first areas where people think about mathematics being applied are in the sciences and engineering. Yet mathematics plays a large role in the efficiency of sports. Coaches constantly try to find ways to get the most out of their athletes, and sometimes they turn to mathematics for help. This help may include the best batting order for a team to maximize the number of runs it can score or the putting together of a program for an Olympic skater so that the jumps the skater makes take advantage of the scoring bonus when these jumps are performed later in a program when tiredness starts to set in. There are also mathematical issues involved in scoring systems for some of the complex and subjective aspects of scoring sports events.

However, the sheer magnitude of the number of games that must be played in league sports creates a large domain for mathematics to assist in the efficient operation of sports. This runs the gamut from "intellectual" sports such as bridge, whist, and chess, to sports such as baseball, football, basketball, soccer, and cricket. Here I will limit myself to some of the fascinating mathematics of sports scheduling and some related fairness and optimization questions that use relatively elementary or quick starting methods.

One way of getting insight into a complex environment is to classify what one sees and study the objects in each of the categories separately as a way of simplifying things. In fact there are many types of tournaments:

* round robin tournament (each team plays exactly k games against every other team (player))

Note: Very often the value of k = 1 is so each team or player gets to play exactly one game (match) against every other team or player.

* elimination tournament (the tournament progresses n rounds where in each round some players are eliminated and the surviving players are paired in future rounds, where again losers are eliminated)

Note: there are variants of this, especially double elimination tournaments. In this idea losers in the various rounds of elimination play against each other and, thus, a later series of victories can lead to a final victory.

* king of the hill (a player stays on the court for as long as he/she can beat the next challenger)

Perhaps the very first question that arises in scheduling is to design the matches that must occur for a round robin tournament. In a single round robin tournament (SRRT) each team must play exactly one game against every other team. We will devote what follows mostly to single round robin tournaments. Many questions arise where mathematics provides insight. First, there is the issue of scheduling. If there are 8 teams, what is an efficient way to schedule the matches that must take place? Another question, about which there is a huge literature but which will not be treated here, is how to decide on the winner based on the results or scores that the players attain. For example, if one has 8 teams, could the number of wins of the eight teams in decreasing order be 6, 5, 5, 4, 4, 2, 2, 0? Questions about rankings for teams in tournaments are closely related to the issues of ranking candidates in an election or ranking choices for economic policy.

Graph theory helps schedule tournaments

Graph theory, a branch of combinatorics which draws heavily on geometrical ideas, uses diagrams consisting of dots and lines to help get insight into a variety of mathematical problems. The complete graph on n vertices has exactly one edge between every pair of vertices. These graphs are denoted Kn Figure 1 shows K4 and Figure 2 shows K5.

In each case the vertices of the graph are labeled with the names of the people or teams involved in the "tournament" or competition. Think of the vertices (dots) of a complete graph as representing the teams in a tournament and think of an edge joining two teams as being a match played by those two teams. The number of edges of the complete graph with n vertices is n(n-1)/2, which is the number of matches that must be carried out in order to have each team play every other team exactly once. Note that in the graph Kn each vertex has n-1 edges at each vertex. The number of edges at a vertex of a graph is known as its degree or valence.

Consider first the case where there are 4 teams that must play each other. This means a total of 4(3)/2 = 6 matches must be played. These matches could be played in 6 time slots, say one a week for 6 weeks. However, it might be desirable if venues (rooms playing fields) for the matches are available to have several matches per time slot and the games be completed over a shorter period of time. Thus, since there are 4 players, and 4/2 is 2, we could consider having two matches per time slot, and complete the tournament in three weeks rather than 6 weeks. When I use the phase "time slot," there are various possibilities as to how the matches are actually played. Note that two matches per time slot might mean that there would be two games at exactly the same time or that the games be played in the morning and afternoon on the same "court" of a single day. There are a variety of terms used other than time slots, and a common one is "rounds," which I will use interchangeably with time slots and Event Window. Figure 3 shows the details of how the scheduling could work.

Edges in the graph that have the same color would occur during one time slot. Thus, for Event Window 1 (shown in blue) there would be matches between team 0 and team 3 and team 1 and team 2 for Event Window 2 (shown in black) we would pair team 0 and team 1 and team 2 and team 3 and for Event Window 3 (shown in red) we would pair team 0 and team 2 and team 1 and team 3.

In attempting to use the ideas above we come to a complication when we try to extend what we have done from 4 teams to 5 teams. Since 5 is an odd number we can not merely have all the teams play in pairs during an Event Window. There is a natural way to handle this problem. The concept of a "bye" in sports scheduling refers to a team's not having to play a match (game) during a particular Event Window. If one has 5 teams, there are 10 matches (games) that must be carried out for a round robin tournament where each team plays every other. Since 5/2 is not an integer, we can not play 3 games per Event Window but we can play 2 games per Event Window (4 teams play) and assign a bye to one team. Thus, in five Event Windows we can schedule the whole tournament. You can see the way a schedule for the five Event Windows can be constructed and see the team which has a bye in each Event Window by consulting Figure 4.

The edges in different colors signify which teams play in an Event Window. For example, the two yellow edges tell one can have teams 0 and 3 and 1 and 2 play each other in a single Event Window for that event window team 4 would get a bye. The other pairings for each Event Window can be similarly handled. Note that there is considerable flexibility in the arrangement of the colors into the five Event Windows.

If a collection of edges are disjoint from each other it is called a matching. If a graph G has a matching M which includes all of the vertices of the graph, then M is said to be a perfect matching. A necessary condition for a perfect matching is that the number of vertices of the graph be even. K4 has a perfect matching while K5 does not. However, it is not difficult to find examples, such as the one in Figure 5, which has an even number of vertices, every vertex of valence 3 (e.g. 3 edges at a vertex), but for which there is no perfect matching.

A pioneer in using graph theory as a tool for solving scheduling problems has been Dominique De Werra, who has spent much of his career at the Polytechnic University of Lausanne.

During that time he has made a variety of contributions to sports scheduling and operations research in general. Many of the basic results were described during the 1980's by De Werra. Interest in this subject has grown so large that there is now an online discussion group devoted to sports scheduling issues from both practical and theoretical viewpoints.

Another name for a perfect matching is a 1-factor. A k-factor one is a subgraph of the graph which includes all the vertices of the graph and where each vertex in the subgraph has the valence (degree) k. So when a graph has a 1-factor, we can think of the vertices as teams and the edges as games which the vertices (teams) joined by an edge play against each other. Returning to our sports scheduling situation, when we have a complete graph which has an even number of vertices, we can ask if it has a collection of 1-factors which include all of the edges of the graph. The coloring we found for K4 in Figure 3 shows that this graph has a 1-factorization into three 1-factors. Because of the special way we drew K4 it may not be clear that we can continue to find 1-factorizations of complete graphs with even numbers of vertices. To see the different in suggestiveness of different drawings, look at this drawing of K4 (Figure 6).

In this version we can see that the edges of different colors can be interpreted as being in "parallel classes." Even though the edges 02 and 13, which are black, appear to meet, they meet at a point which is not a vertex so we will think of this drawing as having three parallel classes. One can, in fact, interpret this diagram as a finite affine plane with 4 points. Every line of the plane has two points on it. There are six lines and 3 lines through every point.

Now we move up to round robin tournaments with 6 teams (Figure 7). Fifteen matches are to be played. Since there are 6 players this means having 3 games per Event Window for 15/3 = 5 Event Windows.

In light of what happened for four teams it is tempting to take a boundary edge 01 in Figure 7 of the regular hexagon shown, and construct a matching by using the edges that do not meet this edge (that are parallel to it, as it were). If we do this we get the games: 01, 25, and 34. Proceeding around the boundary we get another two groups of matches: 12, 03, 45, and 23, 14, 05. This seems to take us off to a good start. There are 6 edges remaining so our hope is to group these into two sets of size 3. However, unfortunately the six edges that remain form two disjoint triangles: edges 02, 24, 04 and 13, 15, 35. Now since we can not pick two disjoint edges from either of these triangles we reach a dead end. There is no way we can take our initial group of teams for the first three time windows and extend the result to two more time windows! Although mathematicians love to reason by analogy and try to apply simple principles to solve a problem at hand, sometimes the analogy may not hold up, as we see in this case.

However, we will not lose heart. Perhaps we can try some alternate systematic way to schedule 6 teams in a round robin tournament. Here is a method that works and generalizes. Here we number the teams from 1 to n rather than from 0 to n-1. We will consider only the case with an even number of teams, since when there is an odd number of teams, as already explained we can add a fictional team and whenever a real team is asked to play the fictional team, the real team has a bye.

Consider the case with 6 teams. Construct an initial table with the first half of the teams listed consecutively in the first row and the last half of the teams listed in reverse order in the next row. The teams that line up in the table will play in the first round.

We can use the graph below to visualize what is going on. The edge 1 to 6 is shown vertically in a diagram where the vertex 1 is placed at the "center" of a regular pentagon and the numbers 2 and 3 are listed clockwise starting from 1, while the numbers 5 and 4 are shown counterclockwise starting at 6. In the diagram shown the edges which make up the sides of the regular convex pentagon are omitted. Only the edges which make up the pairings in one round for the teams are shown. The other edges of the matching (in addition to the pairing of 1 and 6, shown in red) are shown horizontally in blue. (Two colors are used to highlight the different role of the vertical and horizontal edges in the diagram. However, in the partition of the complete graph on 6 vertices into 3 disjoint edges, these three edges would be in one color class.) Since there are 5 rounds each with 3 edges we can account for all of the 15 edges in the complete graph on 6 vertices.

To get to the pairings for the next round we will fix the first team in the cell in the first row and first column but think of all the other teams as being in a necklace on a piece of string listed in the clockwise direction: 2, 3, 4, 5, and 6. Now rotate the necklace one position clockwise and record the entries into a new table: 6, 2, 3, 4, 5. We were thinking of 2 as being at the top of the clock, so after the rotation the last entry in the list is now at the top of the clock.

Here is an analogue of the diagram above after the rotation of the vertices in the regular pentagon:

We repeat this rotation operation until we exhaust the new pairings and return to the start.

with a corresponding labeled graph:

Another rotation would bring us back to where we started, so we are done.

Here are the first few founds of the schedule for 8 teams. Can you find the teams that play in the remaining rounds?

and a graph-theoretical way to see what is going on:

One additional sample of the rotation in the visual graph theory environment:

Earlier we pointed out the problem of an odd number of teams. The way out was to use a bye in each Event Window. Is there a way to use the elegant system above to unify the odd versus even case treatment? Mathematicians like the idea of taking an elegant way of handling something to deal with a related but "annoying" situation. The way to do this in the current situation is to always work with an even number of teams or players, even when the actual number of players is odd. When we have an odd number of players, we will always think of player number 1 as a "fictional" or nonexistent team. Now, whenever a player is assigned the fictional player 1 as an opponent, that real player is given a bye. Since, for example, in the pairings for 6 teams with the real players 2 to 6, in the diagrams above the red pairings with the fictional team 1 appear exactly once in the order 6, 5, 4, 3, 2. Thus, player 4 has a bye in the third round.

A common concern of mathematics is seeing things to be the same or equivalent when considered from some point of view. For example, the fractions 3/6 and 1/2 certainly look different but they can be used interchangeably in calculations which involve fractions because they are "equivalent" rational numbers. One might wonder if the patterns of scheduling tournaments that are derived from 1-factorizations of complete graphs are equivalent or different. Not surprisingly, if one has two teams there is only one way to schedule a tournament between them. What about 4-player tournaments such as we constructed a schedule for based on the edge colorings of the complete graph on 4 vertices in Figure 1? There we named the players 0 to 3 but in the "rotation" procedure shown above we used the names 1 to 4 for the players. Clearly, renaming of the players does not change a schedule. However, one might be interested in how many essentially different ways there are of scheduling 2n teams. First done by hand and then continuing with computers this question is being attacked. It is tempting to be lulled into "complacency" by the slow start of this sequence: for two players there is 1 schedule, for 4 players and 6 players 1 schedule, and for 8 players only 6 are possible. However, things take off from here as shown by the sequence of values starting at n = 2 (n even):

1, 1, 1, 6, 396, 526915620, 1132835421602062347

This shows that one can try to achieve some other goals by selecting schedules that might have other "desirable" properties beyond the easy way of describing the construction that we have indicated above.

Extensions and generalizations

One natural pragmatic concern for scheduling tournaments is where the games are played. For some tournaments the games may be played on "neutral" territory where the opponents are not at some advantage because of being on their home field or having the encouragement of the hometown fans. However, in a lot of sports there is this issue of home and away games. Is there some easy way to take this issue into account?

The diagram below (Figure 15) shows a copy of the 3 matchings of K4 where each edge has been assigned an orientation, that is, an arrow head has been placed on each edge. I will use the "standard convention" from the scheduling literature that a directed edge from i to j will mean that for the match between i and j, the game will be a home game for j. Using H for home and A for away, the schedule shown with the arrows gives rise to this pattern of rounds with away and home games indicated:

Round 1 (Red matching): 0 (away) plays 2 (home) 1 (home) plays 3 (away)

Round 2 (Black matching): 0 (away) plays 1 (home) 2 (home) plays 3 (away)

Round 3 (Blue matching): 0 (home) plays 3 (away) 1 (home ) plays 2 (away).

One can record this information in a somewhat different way. For each player, list in the three rounds whether its games are home or away.

Team 0: A (Red), A (Black), H (Blue)

Team 1: H (Red), H (Black), H (Blue)

Team 2: H (Red), H (Black), A (Blue)

Team 3: A (Red), A (Black), A (Blue)

This does not seem quite fair because team 1 plays all three of its games at home while team 3 plays all three of its games away. However, since there are 4 teams and 6 matches we can not equalize the number of home and away games for each team. Put differently, since there are three rounds, in each round the number of home and away games can not be equal for any team!

Could one at least find a schedule where the number of home and away games differ in absolute value by 1?

The answer is that for a four-player tournament such a schedule is possible.

Round 1 (Red matching): 0 (away) plays 2 (home) 1 (home) plays 3 (away)

Round 2 (Black matching): 0 (home) plays 1 (away) 2 (home) plays 3 (away)

Round 3 (Blue matching): 0 (away) plays 3 (home) 1 (home ) plays 2 (away).

and in terms of the patterns for each team:

Team 0: A (Red), H (Black), A (Blue)

Team 1: H (Red), A (Black), H (Blue)

Team 2: H (Red), H (Black), A (Blue)

Team 3: A (Red), A (Black), H (Blue)

In considering the pattern of home and away games one might want to have home and away games alternate for each team or, from a different point of view all the home and away games be in a consecutive block. When it might be possible, one might also want to have equal numbers of home and away games. Now, for situations with an even number of teams since the degree (number of edges at a vertex) of a vertex in the complete graph is odd, we can not have equal numbers of home and away games. For complete graphs with an odd number of vertices we can ask for the even number of games each team plays to be equally divided between home and away games, but we have to recall that in each round there will exactly one bye if all the other teams play in that round.

What are the tools that mathematics has brought to bear on these scheduling questions? In the graph theory realm the methods involve finding 1-factorizations of complete graphs, as well as other ways to partition (organize the edges into disjoint subsets) the edges of complete graphs. Sometimes one can a find a 2-factor (a subgraph where each edge has degree (valence) 2) and a 1-factor which together account for all the edges of a complete graph. For example, in Figure 3 if one takes the union of the edges with any two different colors, we get a 2-factor of the graph, and edges of the third color form a 1-factor. In fact, this idea can be used to get the appealing way to orient the edges of the complete graph on four vertices shown in Figure 16. If we orient the edges of the 2-factor to form a "directed cycle," then we can orient the remaining two edges to get a home-away pattern that is symmetrical for a tournament with 4 teams. (This pattern does not have totally equal treatment for the four teams, but if there is a schedule of games for the fall and another for the spring the roles of the teams can be interchanged and fairness restored over the longer period.)

Mathematics has responded to the need for finding "good" schedules in the most commonly required settings with a variety of ideas. If, in light of the discussion above, one limits oneself to scheduling an even number n of teams (players), then one can carry out two phases, in an attempt at producing a "nice" schedule:

Schedule n/2 matches for each of the n-1 rounds (Event Windows).

What is produced by this phase is determining for each team which opponent it will have in each of the rounds.

For each of the matches scheduled in Phase I between pairs of players (teams) one decides which of the two teams in the match plays at home and which team plays away.

For the "opponent schedule" produced in phase I, phase II produces a "home-away assignment."

In some situations the issue of home/away does not matter but it usually does. Thus, for each team, one can produce a sequence of n-1 H's and A's which represent the home/away pattern of games that team must play.
For example, for the opponent schedule generated from Figure 16 we have for Team 0 the sequence AHA and for Team 3 the sequence AAH. It is commonly viewed that consecutive games at home or away are undesirable. In the scheduling literature two such consecutive home or away games is called a break. It turns out to be a classical insight from mathematical analysis that any home-away assignment for an "opponent schedule" can not be done with fewer than n-2 breaks.

The proof of this fact is rather appealing, so here goes. First, no two teams can have identical home-away patterns because if they did, they would not get to play each other, which is required in a round robin tournament! So we can have at most one home-away sequence where home and away alternate and which starts and ends with home, and at most one such home away sequence that starts and ends with away. Thus, we must have at least n-2 teams which have at least one break!

The opponent schedule given in Figure 3 and the home-away schedule derived from this opponent schedule as in Figure 16 has exactly 4-2 =2 breaks, and, thus, achieves the minimum. If one is given an opponent schedule S one can let Bminimum denote the minimum number of breaks taking into account all ways of assigning home and away patterns to the paired teams in S. It is known (Dominique de Werra showed this) that one can (for even n) find an opponent schedules where Bminimum is n-2. It has also been shown that for a given opponent schedule one can decide whether or not it can be given a home-away assignment (that is, one can carry out Phase II, above) that achieves n-2 breaks in polynomial time. Yet, given an arbitrary S the computational complexity of finding the value of Bminimum is unclear. It may require exponential time to carry out this calculation.

Our discussion has given some clues to the many ways that graph theory can be used to get insights into scheduling problems for different sports. Another area of mathematics that has also very fruitful for this purpose is the theory of block designs. Many web sites now offer the opportunity to print out tournament arrangements for various numbers of teams.

When you sit down to watch your favorite sports star or team I hope you will recognize the behind-the-scenes role that mathematics is playing in bringing these events to you and making it possible to have fair, competitive and efficient sports events. If you are a "little league" mom or dad or participant, perhaps you can enjoy the mathematics behind the sports, as well as the sport itself.


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Briskorn, D. and S Knust, Constructing fair sports league schedules with regard to strength groups, Discrete Applied Mathematics 158 (2010) 123-135.

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Post, G. and G. Woeginger, Sports tournaments, home-away assignments, and the break minimization problem, Discrete Optimization, 3 (2006) 165-173.

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Rosa, A. and W. Wallis, Premature sets of 1-factors or how not to schedule round-robin tournaments, Discrete Applied Mathematics, 4 (1982) 291-297.

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Smith, M. and J. Mather, K.. Gibson, S. Jowett, Measuring the dynamic response of a golf club during swing and impact, Photogrammetric Record, 16 (1998), 249-257.

Trick, M. A schedule-then-break approach to sports timetabling, in Proceedings of the 3rd Conference on Practice and Theory of Automated Timetabling, PATAT' 2001, Lecture Notes in Computer Science, Volume 2079, Springer, Berlin, 2001, pp. 242-253.

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Those who can access JSTOR can find some of the papers mentioned above there. For those with access, the American Mathematical Society's MathSciNet can be used to get additional bibliographic information and reviews of some these materials. Some of the items above can be accessed via the ACM Portal , which also provides bibliographic services.

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AP Calculus AB is an introductory college-level calculus course. Students cultivate their understanding of differential and integral calculus through engaging with real-world problems represented graphically, numerically, analytically, and verbally and using definitions and theorems to build arguments and justify conclusions as they explore concepts like change, limits, and the analysis of functions.

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Based on the Understanding by Design® (Wiggins and McTighe) model, this course framework provides a clear and detailed description of the course requirements necessary for student success. The framework specifies what students must know, be able to do, and understand, with a focus on big ideas that encompass core principles, theories, and processes of the discipline. The framework also encourages instruction that prepares students for advanced coursework in mathematics or other fields engaged in modeling change (e.g., pure sciences, engineering, or economics) and for creating useful, reasonable solutions to problems encountered in an ever-changing world.

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Simplify the expression below. Which of the following is correct?

Solve the equation below. Which of the following is correct?

The correct answer is (D). First divide both sides by 3:

The above is known as what type of graph?

Simplify the expression below.

Which of the following is correct?

The correct answer is (B). The “FOIL method” is the easiest way to remember how to multiply two-termed expressions. Multiply the First two terms, then the Outer two terms, then the Inner two terms, and then the Last two terms, then sum all four to arrive at the answer:

Timmy can usually make six paper airplanes in an hour. However, if he gets interrupted by his parents, he can only make four per hour. His friend John can make seven paper airplanes per hour.

One day, Timmy and John decide to have a three-hour long contest to see who can make the most paper airplanes. During the contest, Timmy is interrupted once every hour. John also had to take a break and do chores for an hour. How many more paper airplanes does the winner make?

If a car travels 360 kilometers in 5 hours, how far will it travel in 9 hours?

The correct answer is (D). Given that it took 5 hours to travel 360km, we can set up a ratio equation to figure out how far the car will travel in 9 hours:

What degrees Fahrenheit is it at 30 degrees Celsius?

The correct answer is (A). Plug in °C = 30 into the given formula and solve for °F:

A toy manufacturer makes 15,000 toys a year. The company randomly selects 300 of the toys to sample for inspection. The company discovers that there are 5 faulty toys in the sample. Based on the sample, how many of the 15,000 total toys are likely to be faulty?

The correct answer is (B). The sample indicates that 5 out of every 300 randomly selected toys will be faulty. Consequently, a proportion can be set up that relates the unknown number of faulty toys in the total number of toys to the ratio of faulty toys to the sample:

$T$ is the unknown number of faulty toys in the total. Multiply both sides by 15,000 and then divide the left side by 300 to solve for $T$:

At a comic book store, Robert purchased three comics for $2.65 each. If he paid with a $20 bill, how much change did he receive?

The correct answer is (B). Three comics at $2.65 would equal a total of $7.95. The change would equal:

Simplify the expression below.

Which of the following is correct?

  1. Multiply the whole number by the denominator.
  2. Add the answer from step 1 to the numerator.
  3. Write the answer from step 2 above the denomintor.

To mulitply fractions simply multiply across the top and the bottom:

Express $dfrac$ as its closest rounded percentage.

Amanda wants to paint the walls of a room. She knows that each can of paint contains one gallon. A half gallon will cover a 55 square feet of wall. Each of the four walls is 10 feet high. Two of the walls are 10 feet wide and two of the walls are 15 feet wide. How many 1-gallon buckets of paint does Amanda need to buy in order to fully paint the room?

The correct answer is (B). First, we must calculate the area of the walls Amanda wants to paint. Two of the walls are 10 x 10 and two of the walls are 10 x 15:
2 (10 x 10) = 200
2 (10 x 15) = 300
So the total square footage of the walls is 500.

If a half gallon of paint will cover 55 square feet, then each gallon will cover 110 square feet. Four gallons would only cover 440 square feet. Five gallons will cover 550 square feet, which will be enough for all 500 square feet of walls.

You’re Gonna Love 28 Days of STEM Activities and STEAM Activities for Kids

This year we’re doing the weekly topics a little bit differently… I hear a lot from readers asking for a tech project or an art project for their kids or students. So this year we’re going to round-up some amazing projects with a focus on each of the buckets in STEM & STEAM. But don’t worry, they’re still going to be integrated projects that check off more than one bucket!

The distance and midpoint formulas

We want to calculate the distance between the two points (-2, 1) and (4, 3). We could see the line drawn between these two points is the hypotenuse of a right triangle. The legs of this triangle would be parallel to the axes which mean that we can measure the length of the legs easily.

We'll get the length of the distance d by using the Pythagorean Theorem

This method can be used to determine the distance between any two points in a coordinate plane and is summarized in the distance formula

The point that is at the same distance from two points A (x1, y1) and B (x2, y2) on a line is called the midpoint. You calculate the midpoint using the midpoint formula

We can use the example above to illustrate this

$ m =left ( frac<4+(-2)> <2> ight ),: : left ( frac<3+1> <2> ight )=$