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Week 2
Due for this week…
Homework 2 (on MyMathLab – via the Materials Link) Monday night at 6pm.
Read Chapter 3 and 8.1, 8.2 Do the MyMathLab Self-Check for week 2. Learning team planning for week 5.
Slide 2Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Introduction to Equations
Basic Concepts
The Addition Property of Equality
The Multiplication Property of Equality
2.1
The Linear Equation
It has one variable : x
ax + b = 0
Where a and b are real numbers and a is not equal to 0.
A photo album of linear equations
2x+3=0 (say cheese!) x+8=0 3x=7 2x+5=9-5x 3+5(x-1)=-7+x 3=8x+5 Etc.
How to legally play with linear equations.
Most of the magic we work with are really obvious tricks. Remember what happens if we multiply by things
that equal 1? e.g. 5/5 10/10 ?
Adding the SAME THING to both sides of an equation does not change the equation…
The Addition Property of Equality
So if you start with a = b You don’t change anything if you take a new
number or variable and add it to both sides!
a+c = b+c
The GOAL
We want x all by itself on one side and the rest of the junk on the other (so we can use our calculator or fingers).
The technical term for this is: solving for x
Slide 9Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 10Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE
Solution
Using the addition property of equality
Solve each equation.
a. x + 21 = 9 b. n – 8 = 17
b. a. x + 21 = 9
x + 21 + (−21) = 9 + (−21)
x + 0 = −12
x = −12
n – 8 = 17
n – 8 + 8 = 17 + 8
n = 25
n + 0 = 25
The solution is −12. The solution is 25.
** Try exercises 17-26
Slide 11Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE
Solution
Solving and checking a solution
Solve the equation −7 + x = 12 and then check the solution.
Check:
Isolate x by adding 7 to each side.
−7 + x = 12
−7 + 7 + x = 12 + 7
0 + x = 19
x = 19
12 = 12The solution is 19.
The answer checks.
−7 + x = 12
−7 + 19 = 12?
** Try exercises 27-34
Slide 12Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 13Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE
Solution
Using the multiplication property of equality
Solve each equation.
a. b. −11a = 33
b. a.
18
6x
18
6x
8 61
66 x
1 48x
48x
−11a = 33
11 1
3
1
11 3a
3a
The solution is 48.
The solution is −3.
** Try exercises 41-48
Slide 14Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley-58
EXAMPLE
Solution
Solving and checking a solution
Solve the equation and then check the solution.
Check:
The solution isThe answer checks.
2 2
5 9x
2 2
5 9x
9 2 2 9
2 5 9 2x
9
15
x
9
5x
9.
5
2 2
5 9x
? 9
5
2 2
5 9
2 2
5 5
** Try exercises 49-58
Slide 15Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE
Solution
Application
A veterinary assistant holds a cat and steps on a scale. The scale reads 153 lbs. Alone the assistant weighs 146 lbs.
a. Write a formula to show the relationship of the weight of the cat, x, and the assistant.
b. Determine how much the cat weighs.
a. 146 + x = 153 146 + x = 153
146 + (−146) + x = 153 − 146
b.
x = 7
The cat weighs 7 lbs.** Try exercises 61-70
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Linear Equations
Basic Concepts
Solving Linear Equations
Applying the Distributive Property
Clearing Fractions and Decimals
Equations with No Solutions or Infinitely Many Solutions
2.2
Slide 17Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
If an equation is linear, writing it in the form ax + b = 0 should not require any properties or processes other than the following:
• using the distributive property to clear parentheses,• combining like terms,• applying the addition property of equality.
Slide 18Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE
Solution
Determine whether an equation is linear
Determine whether the equation is linear. If the equation is linear, give values for a and b.
a. 9x + 7 = 0 b. 5x3 + 9 = 0 c.
b. This equation is not linear because it cannot be written in the form ax + b = 0. The variable has an exponent other than 1.
a. The equation is linear because it is in the form ax + b = 0 with a = 9 and b = 7.
57 0
x
c. This equation is not linear because it cannot be written in the form ax + b = 0. The variable appears in the denominator of a fraction. ** Try exercises 13-26
Slide 19Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE
Solution
Using a table to solve an equation
Complete the table for the given values of x. Then solve the equation 4x – 6 = −2.
From the table 4x – 6 = −2 when x = 1. Thus the solution to 4x – 6 = −2 is 1.
x −3 −2 −1 0 1 2 3
4x − 6 −18
x −3 −2 −1 0 1 2 3
4x − 6 −18 −14 −10 −6 −2 2 6
** Try exercises 27-30
Slide 20
Rules for Happiness
First do whatever adding and subtracting you can do…
THEN do whatever multiplication and division you can do.
THEN you should be done!
Slide 22Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE
Solution
Solving linear equations
Solve each linear equation.
a. 12x − 15 = 0 b. 3x + 19 = 5x + 5
b. a. 12x − 15 = 0
12x − 15 + 15 = 0 + 15
12x = 15 12 15
12 12
x
15
12x
5
4
3x + 19 = 5x + 5
3x − 3x + 19 = 5x − 3x + 5
19 = 2x + 5
19 − 5 = 2x + 5 − 5
14 = 2x 14
2 2
2x
7 x** Try exercises 39-42
Slide 23Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE
Solution
Applying the distributive property
Solve the linear equation. Check your solution. x + 5 (x – 1) = 11
6x − 5 = 11
16
6x
6 16
6 6
x
x + 5 (x – 1) = 11 x + 5 x – 5 = 11
6x − 5 + 5 = 11 + 5 6x = 16
8
3
Slide 24Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE continued
Checkx + 5 (x – 1) = 11
8 85 1 11
3 3
8 8 35 11
3 3 3
8 55 11
3 3
8 2511
3 3
3311
3
11 11
The answer checks, the solution is
8.
3
** Try exercises 43-52
Slide 25Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE
Solution
Clearing fractions from linear equations
Solve the linear equation. 1 1
52 6
x x
1 156 6
2 6x x
3 30x x
2 30x
15x
1 15
2 6x x
The solution is 15.
** Try exercises 55-58
Slide 26Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE
Solution
Clearing decimals from linear equations
Solve the linear equation. 5.3 0.8 7x
5.3 0.8 7x 5.3 0.10 18 7 0x
53 8 70x 53 8 753 0 53x
8 17x 8 17
8 8
x
17
8x
The solution is17
.8
** Try exercises 53-54
Equations with No Solutions or Infinitely Many Solutions
An equation that is always true is called an identity and an equation that is always false is called a contradiction.
Slide 27
Slide 28Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE
Solution
Determining numbers of solutions
Determine whether the equation has no solutions, one solution, or infinitely many solutions.
a. 10 – 8x = 2(5 – 4x)b. 7x = 9x + 2(12 – x) c. 6x = 4(x + 5)
a. 10 – 8x = 2(5 – 4x)
10 – 8x = 10 – 8x
– 8x = – 8x
0 = 0
Because the equation 0 = 0 is always true, it is an identity and there are infinitely many solutions.
Slide 29Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE continued
b. 7x = 9x + 2(12 – x)
Because the equation 0 = 24 is always false, it is a contradiction and there are no solutions.
c. 6x = 4(x + 5)
7x = 9x + 24 – 2x
7x = 7x + 24
0 = 24
6x = 4x + 20
2x = 20
x = 10
Thus there is one solution.
** Try exercises 63-72
Slide 30
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Introduction to Problem Solving
Steps for Solving a Problem
Percent Problems
Distance Problems
Other Types of Problems
2.3
Slide 32Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 33Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE
Solution
Translating sentences into equations
Translate the sentence into an equation using the variable x. Then solve the resulting equation.
a. Six times a number plus 7 is equal to 25.b. The sum of one-third of a number and 9 is 18.c. Twenty is 8 less than twice a number.
a. 6x + 7 = 25
6x = 186 18
6 6
x
3x
b. 19 18
3x
27x
19
3x
c. 20 = 2x − 8
28 = 2x
14 = x
28 2
2 2
x
** Try exercises 11-18
Slide 34Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE
Solution
Solving a number problem
The sum of three consecutive integers is 126. Find the three numbers.
Step 1: Assign a variable to an unknown quantity.
n + (n + 1) + (n + 2) = 126
n + 1: next integer
n + 2: largest integer
Step 2: Write an equation that relates these unknown quantities.
n: smallest of the three integers
Slide 35Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE continued
Step 3: Solve the equation in Step 2.
n + (n + 1) + (n + 2) = 126
Step 4: Check your answer. The sum of these integers is 41 + 42 + 43 = 126.
The answer checks.
(n + n + n) + (1 + 2) = 126
3n + 3 = 1263n = 123n = 41
So the numbers are 41, 42, and 43.
** Try exercises 19-28
Note: To write x% as a decimal number, move the decimal point in the number x two places to the left and then remove the % symbol.
Slide 36
Slide 37Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE
Solution
Converting percent notation
Convert each percentage to fraction and decimal notation.
a. 47% b. 9.8% c. 0.9%
a. Fraction Notation:47
47% .100
Decimal Notation: 47% 0.47.
b. Fraction Notation:9.8 98 49 2 49
9.8% .100 1000 500 2 500
Decimal Notation: 9.8% 0.098.
c. Fraction Notation:0.9 9
0.9%100 1000
Decimal Notation: 0.9% 0.009. ** Try exercises 35-42
Slide 38Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE
Solution
Converting to percent notation
Convert each real number to a percentage.
a. 0.761 b. c. 6.3
a. Move the decimal point two places to the right and then insert the % symbol to obtain 0.761 = 76.1%
2
5
b. 2 2
0.40, so 40%.5 5
c. Move the decimal point two places to the right and then insert the % symbol to obtain 6.3 = 630%. Note that percentages can be greater than 100%.
** Try exercises 43-54
Slide 39Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE
Solution
Calculating a percent increase
The price of an oil change for an automobile increased from $15 to $24. Calculate the percent increase.
old value
ol
-
d
new
valu
e
value100 15
1
24
5
- 100 60%
** Try exercises 55-56
Slide 40Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE
Solution
Solving a percent problem
A car salesman sells a total of 85 cars in the first and second quarter of the year. In the second quarter, he had an increase of 240% over the previous quarter. How many cars did the salesman sell in the first quarter?
Step 1: Assign a variable.
x: the amount sold in the first quarter.
Step 2: Write an equation.
x + 2.4x = 85
Slide 41Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE continued
Step 3: Solve the equation in Step 2.
Step 4: Check your answer. An increase of 240% of 25 is 2.4 × 25 = 60.
Thus the amount of cars sold in the second quarter would be 25 + 60 = 85.
In the first quarter the salesman sold 25 cars.
x + 2.4x = 85
3.4x = 85
x = 25
** Try exercises 57-60
Slide 42Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE
Solution
Solving a distance problem
A truck driver travels for 4 hours and 30 minutes at a constant speed and travels 252 miles. Find the speed of the truck in miles per hour.
Step 1: Let r represent the truck’s rate, or speed, in miles.
Step 2: The rate is to be given in miles per hour, so change the 4 hours and 30 minutes to 4.5 or 9/2 hours.
9252
2r
d = rt
Slide 43Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE continued
Step 3: Solve the equation.9
2522
r
2 2
9 9
9252
2r
56 r
The speed of the truck is 56 miles per hour.
Step 4:
d = rt9
56 252 miles2
The answer checks.** Try exercises 71-74
Slide 44Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE
Solution
Mixing chemicals
A chemist mixes 100 mL of a 28% solution of alcohol with another sample of 40% alcohol solution to obtain a sample containing 36% alcohol. How much of the 40% alcohol was used?
Concentration Solution Amount (milliliters)
Pure alcohol
0.28 100 28
0.40 x 0.4x
0.36 x + 100 0.36x + 36
Step 1: Assign a variable. x: milliliters of 40%
x + 100: milliliters of 36%
Step 2: Write an equation.
0.28(100) + 0.4x = 0.36(x + 100)
Slide 45Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE continued
Step 3: Solve the equation in Step 2.
200 mL of 40% alcohol solution was added to the 100 mL of the 28% solution.
0.28(100) + 0.4x = 0.36(x + 100)
28(100) + 40x = 36(x + 100)2800 + 40x = 36x + 36002800 + 4x = 3600
4x = 800 x = 200
Slide 46Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE continued
Step 4: Check your answer. If 200 mL of 40% solution are added to the 100 mL of 28% solution, there will be 300 mL of solution.
200(0.4) + 100(0.28) = 80 + 28 = 108 of pure alcohol.
The concentration is or 36%. 108
0.36,300
** Try exercises 79-80
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Formulas
Basic Concepts
Formulas from Geometry
Solving for a Variable
Other Formulas
2.4
Slide 48Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE
Solution
Calculating mileage of a trip
A tourist starts a trip with a full tank of gas and an odometer that reads 59,478 miles. At the end of the trip, it takes 8.6 gallons of gas to fill the tank, and the odometer reads 59,715 miles. Find the gas mileage for the car.
DM
G
The distance traveled is 59,715 – 59, 478 = 237 miles and the number of gallons used is G = 8.6. Thus,
6
237
8. 27.6 miles per gallon.
** Try exercises 81-82
Slide 49Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE
Solution
Calculating area of a region
A residential lot is shown.Find the area of this lot.
RA LWThe area of the rectangle:
205 ft
372 ft
116 ft
372 205RA 76,260 square feetRA
The area of the triangle: 12TA bh
12 116 372TA 21,576 square feetTA
Total area = 76,260 + 21,576 = 97,836 square feet.** Try exercises 19-29
Slide 50Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE
Solution
Finding angles in a triangle
In a triangle, the smaller angles are equal in measure and are one-third of the largest angle. Find the measure of each angle.
Let x represent the measure of each of the two smaller angles. Then the measure of the largest angle is 3x, and the sum of the measures of the three angles is given by
3 180x x x
The measure of the largest angle is 3x, thus 36 ∙ 3 = 108°.The measure of the three angles are 36°, 36°, and 108°.
5 180x
5 180
5 5
x
36x
** Try exercises 31-38
Slide 51Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE
Solution
Finding the volume and surface area of a box
Find the volume and the surface area of the box shown.
V = LHW
The volume of the box is
2 2 2S LW WH LH
12 cm5 cm
6 cm
V = 12 ∙ 6 ∙ 5 V = 360 cm3
The surface area of the box is
2(12)(5) 2(5)(6) 2(12)(6)S 120 60 144S 324 square centimetersS ** Try exercises 41-46
Slide 52Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE
Solution
Calculating the volume of a soup can
A cylindrical soup can has a radius of 2 ½ inches and a height of inches. Find the volume of the can.
2V r h
∙h
r
245
8
5
2V
585
1125
32V
110.45 cubic inchesV
** Try exercises 51
Slide 53Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE
Solution
Solving for a variable
Solve each equation for the indicated variable.
a. b. 3 for 5
y zx z
for np nm nq p
a. 3 5
y zx
15 x y z
15 x y z
15 z x y
b. for np nm nq p np nq nm
( )np n q m ( )n q m
pn
p q m
** Try exercises 53-64
Other Formulas
To calculate a student’s GPA, the number of credits earned with a grade of A, B, C, D, and F must be known. If a, b, c, d, and f represent these credit counts respectively, then
4 3 2.
a b c dGPA
a b c d f
Slide 54
Slide 55Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE
Solution
Calculating a student’s GPA
A student has earned 18 credits of A, 22 credits of B, 8 credits of C and 4 credits of D. Calculate the student’s GPA to the nearest hundredth.
Let a = 18, b = 22, c = 8, d = 4 and f = 0
4 18 3 22 2 8 4
18 22 8 4 0GPA
158
52 3.04
The student’s GPA is 3.04.
** Try exercises 69-72
Slide 56Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE
Solution
Converting temperature
The formula is used to convert degrees Fahrenheit to degrees Celsius.
Use this formula to convert 23°F to an equivalent Celsius temperature.
= −5°C
59 32C F
532
9C F
25
329
3C
59
9C
** Try exercises 73-80
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Linear Inequalities
Solutions and Number Line Graphs
The Addition Property of Inequalities
The Multiplication Property of Inequalities
Applications
2.5
Slide 58Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Solutions and Number Line Graphs
A linear inequality results whenever the equals sign in a linear equation is replaced with any one of the symbols <, ≤, >, or ≥.
x > 5, 3x + 4 < 0, 1 – y ≥ 9
A solution to an inequality is a value of the variable that makes the statement true. The set of all solutions is called the solution set.
Slide 59Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE Graphing inequalities on a number line
Use a number line to graph the solution set to each inequality.
a.
b.
c.
d.
1x
1x
5x
2x
** Try exercises 13-20
Each number line graphed on the previous slide represents an interval of real numbers that corresponds to the solution set to an inequality.Brackets and parentheses can be used to represent the interval. For example:
Slide 60Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Interval Notation
1x (1, )
1x [1, )
Slide 61Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE Writing solution sets in interval notation
Write the solution set to each inequality in interval notation.
a. b.6x 2y
(6, ) ( , 2]
** Try exercises 27-32
Slide 62Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE Checking possible solutions
Determine whether the given value of x is a solution to the inequality. 4 2 8, 7x x
?
?
4 2 <8
4( 87) 2
x
?
28 2 8
?
26 8 e Fals
** Try exercises 33-42
The Addition Property of Inequalities
Slide 63Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 64Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE
Solution
Applying the addition property of inequalities
Solve each inequality. Then graph the solution set.a. x – 2 > 3 b. 4 + 2x ≤ 6 + x
a. x – 2 > 3
x – 2 + 2 > 3 + 2
x > 5
b. 4 + 2x ≤ 6 + x
4 + 2x – x ≤ 6 + x – x
4 + x ≤ 6
4 – 4 + x ≤ 6 – 4
x ≤ 2
** Try exercises 51-58
Slide 65Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
The Multiplication Property of Inequalities
Slide 66Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE
Solution
Applying the multiplication property of inequalities
Solve each inequality. Then graph the solution set.a. 4x > 12 b.
a. 4x > 12 b.
12
4x
4 2
4 4
1x
3x
12
4x
4 (1
( 2)4
4) x
8 x
** Try exercises 59-66
Slide 67Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE
Solution
Applying both properties of inequalities
Solve each inequality. Write the solution set in set-builder notation.a. 4x – 8 > 12 b.
a. 4x – 8 > 12 b.
4 3 4 5x x
4 8 8 12 8x 4 20x
4 3 4 5x x 4 3 3 4 5 3x x x x
4 5x
5x
{ | 5}x x
4 5 5 5x 9 x
{ | 9}x x
** Try exercises 71-100
Slide 68Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Applications
To solve applications involving inequalities, we often have to translate words to mathematical statements.
Slide 69Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE Translating words to inequalities
Translate each phrase to an inequality. Let the variable be x.a. A number that is more than 25.
b. A height that is at least 42 inches.
x > 25
x ≥ 42
** Try exercises 101-108
Slide 70Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE Calculating revenue, cost, and profit
For a snack food company, the cost to produce one case of snacks is $135 plus a one-time fixed cost of $175,000 for research and development. The revenue received from selling one case of snacks is $250.
a. Write a formula that gives the cost C of producing x cases of snacks.
b. Write a formula that gives the revenue R from selling x cases of snacks.
C = 135x + 175,000
R = 250x
Slide 71Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE Calculating revenue, cost, and profit
For a snack food company, the cost to produce one case of snacks is $135 plus a one-time fixed cost of $175,000 for research and development. The revenue received from selling one case of snacks is $250.
c. Profit equals revenue minus cost. Write a formula that calculates the profit P from selling x cases of snacks.
d. How many cases need to be sold to yield a positive profit?
P = R – C= 250x – (135x + 175,000)
= 115x – 175,000
115x – 175,000 > 0 115x > 175,000 x > 1521.74
Must sell at least 1522 cases. ** Try exercises 119-120
End of week 2
You again have the answers to those problems not assigned
Practice is SOOO important in this course. Work as much as you can with MyMathLab, the
materials in the text, and on my Webpage. Do everything you can scrape time up for, first the
hardest topics then the easiest. You are building a skill like typing, skiing, playing a
game, solving puzzles. NEXT TIME: Linear Equations w/2 variables and
Graphing + slope and y-intercepts