Xem mẫu
- – THEA MATH REVIEW –
Then add the 1 pound to the 4 pounds:
4 pounds 25 ounces = 4 pounds + 1 pound 9 ounces = 5 pounds 9 ounces
S UBTRACTION M EASUREMENTS
WITH
1. Subtract like units if possible.
2. If not, regroup units to allow for subtraction.
3. Write the answer in simplest form.
For example, 6 pounds 2 ounces subtracted from 9 pounds 10 ounces.
9 lb 10 oz Subtract ounces from ounces.
– 6 lb 2 oz Then subtract pounds from pounds.
3 lb 8 oz
Sometimes, it is necessary to regroup units when subtracting.
Example
Subtract 3 yards 2 feet from 5 yards 1 foot.
Because 2 feet cannot be taken from 1 foot, regroup 1 yard from the 5 yards and convert the 1 yard
to 3 feet. Add 3 feet to 1 foot. Then subtract feet from feet and yards from yards:
4 4
5 yd 1 ft
– 3 yd 2 ft
1 yd 2ft
5 yards 1 foot – 3 yards 2 feet = 1 yard 2 feet
M ULTIPLICATION M EASUREMENTS
WITH
1. Multiply like units if units are involved.
2. Simplify the answer.
Example
Multiply 5 feet 7 inches by 3.
5 ft 7 in Multiply 7 inches by 3, then multiply 5 feet by 3. Keep the units separate.
3
15 ft 21 in Since 12 inches = 1 foot, simplify 21 inches.
15 ft 21 in = 15 ft + 1 ft 9 in = 16 ft 9 in
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Example
Multiply 9 feet by 4 yards.
First, decide on a common unit: either change the 9 feet to yards, or change the 4 yards to feet. Both
options are explained below:
Option 1:
To change yards to feet, multiply the number of feet in a yard (3) by the number of yards in this
problem (4).
3 feet in a yard 4 yards = 12 feet
Then multiply: 9 feet 12 feet = 108 square feet.
(Note: feet feet = square feet = ft2)
Option 2:
To change feet to yards, divide the number of feet given (9), by the number of feet in a yard (3).
9 feet ÷ 3 feet in a yard = 3 yards
Then multiply 3 yards by 4 yards = 12 square yards.
(Note: yards • yards = square yards = yd2)
D IVISION M EASUREMENTS
WITH
1. Divide into the larger units first.
2. Convert the remainder to the smaller unit.
3. Add the converted remainder to the existing smaller unit if any.
4. Divide into smaller units.
5. Write the answer in simplest form.
Example
Divide 5 quarts 4 ounces by 4.
1. Divide into the larger unit:
1 qt r 1 qt
4 5 qt
– 4 qt
1 qt
2. Convert the remainder:
1 qt = 32 oz
3. Add remainder to original smaller unit:
32 oz + 4 oz = 36 oz
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4. Divide into smaller units:
36 oz ÷ 4 = 9 oz
5. Write the answer in simplest form:
1 qt 9 oz
Metric Measurements
The metric system is an international system of measurement also called the decimal system. Converting units
in the metric system is much easier than converting units in the customary system of measurement. However, mak-
ing conversions between the two systems is much more difficult. The basic units of the metric system are the meter,
gram, and liter. Here is a general idea of how the two systems compare:
Metric System Customary System
1 meter A meter is a little more than a yard; it is equal to about 39 inches
1 gram A gram is a very small unit of weight; there are about 30 grams
in one ounce.
1 liter A liter is a little more than a quart.
Prefixes are attached to the basic metric units listed above to indicate the amount of each unit. For exam-
1
ple, the prefix deci means one-tenth ( 10 ); therefore, one decigram is one-tenth of a gram, and one decimeter is
one-tenth of a meter. The following six prefixes can be used with every metric unit:
Kilo Hecto Deka Deci Centi Milli
(k) (h) (dk) (d) (c) (m)
1 1 1
1,000 100 10 10 100 1,000
Examples
1 hectometer = 1 hm = 100 meters
■
1
■ 1 millimeter = 1 mm =
1,000 meter = .001 meter
■ 1 dekagram = 1 dkg = 10 grams
1
■ 1 centiliter = 1 cL* =
100 liter = .01 liter
■ 1 kilogram = 1 kg = 1,000 grams
1
■ 1 deciliter = 1 dL* =
10 liter = .1 liter
*Notice that liter is abbreviated with a capital letter—L.
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- Metric Prefixes
An easy way to remember the metric prefixes is to remember the mnemonic: “King Henry Died of
Drinking Chocolate Milk”. The first letter of each word represents a corresponding metric heading from
Kilo down to Milli: ‘King’—Kilo, ‘Henry’—Hecto, ‘Died’—Deka, ‘of’—original unit, ‘Drinking’—Deci,
‘Chocolate’—Centi, and ‘Milk’—Milli.
The chart below illustrates some common relationships used in the metric system:
Length Weight Volume
1 km = 1,000 m 1 kg = 1,000 g 1 kL = 1,000 L
1 m = .001 km 1 g = .001 kg 1 L = .001 kL
1 m = 100 cm 1 g = 100 cg 1 L = 100 cL
1 cm = .01 m 1 cg = .01 g 1 cL = .01 L
1 m = 1,000 mm 1 g = 1,000 mg 1 L = 1,000 mL
1 mm = .001 m 1 mg = .001 g 1 mL = .001 L
Conversions within the Metric System
An easy way to do conversions with the metric system is to move the decimal point either to the right or left because
the conversion factor is always ten or a power of ten. Remember, when changing from a large unit to a smaller unit,
multiply. When changing from a small unit to a larger unit, divide.
Making Easy Conversions within the Metric System
When multiplying by a power of ten, move the decimal point to the right, since the number becomes larger. When
dividing by a power of ten, move the decimal point to the left, since the number becomes smaller. (See below.)
To change from a larger unit to a smaller unit, move the decimal point to the right.
→
kilo hecto deka UNIT deci centi milli
←
To change from a smaller unit to a larger unit, move the decimal point to the left.
Example
Change 520 grams to kilograms.
1. Be aware that changing meters to kilometers is going from small units to larger units and, thus,
requires that the decimal point move to the left.
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2. Beginning at the UNIT (for grams), note that the kilo heading is three places away. Therefore, the
decimal point will move three places to the left.
k h dk UNIT d c m
3. Move the decimal point from the end of 520 to the left three places.
520
←
.520
Place the decimal point before the 5: .520
The answer is 520 grams = .520 kilograms.
Example
Ron’s supply truck can hold a total of 20,000 kilograms. If he exceeds that limit, he must buy stabiliz-
ers for the truck that cost $12.80 each. Each stabilizer can hold 100 additional kilograms. If he wants
to pack 22,300,000 grams of supplies, how much money will he have to spend on the stabilizers?
1. First, change 2,300,000 grams to kilograms.
kg hg dkg g dg cg mg
2. Move the decimal point 3 places to the left: 22,300,000 g = 22,300.000 kg = 22,300 kg.
3. Subtract to find the amount over the limit: 22,300 kg – 20,000 kg = 2,300 kg.
4. Because each stabilizer holds 100 kilograms and the supplies exceed the weight limit of the truck by 2,300
kilograms, Ron must purchase 23 stabilizers: 2,300 kg ÷ 100 kg per stabilizer = 23 stabilizers.
5. Each stabilizer costs $12.80, so multiply $12.80 by 23: $12.80 23 = $294.40.
A lgebra
This section will help in mastering algebraic equations by reviewing variables, cross multiplication, algebraic frac-
tions, reciprocal rules, and exponents. Algebra is arithmetic using letters, called variables, in place of numbers.
By using variables, the general relationships among numbers can be easier to see and understand.
Algebra Terminology
A term of a polynomial is an expression that is composed of variables and their exponents, and coefficients. A vari-
able is a letter that represents an unknown number. Variables are frequently used in equations, formulas, and in
mathematical rules to help illustrate numerical relationships. When a number is placed next to a variable, indi-
cating multiplication, the number is said to be the coefficient of the variable.
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Examples
8c 8 is the coefficient to the variable c.
6ab 6 is the coefficient to both variables, a and b.
T HREE K INDS P OLYNOMIALS
OF
Monomials are single terms that are composed of variables and their exponents and a positive or nega-
■
tive coefficient. The following are examples of monomials: x, 5x, –6y3, 10x2y, 7, 0.
Binomials are two non-like monomial terms separated by + or – signs. The following are examples of
■
binomials: x + 2, 3x2 – 5x, –3xy2 + 2xy.
Trinomials are three non-like monomial terms separated by + or – signs. The following are examples of
■
trinomials: x2 + 2x – 1, 3x2 – 5x + 4, –3xy2 + 2xy – 6x.
Monomials, binomials, and trinomials are all examples of polynomials, but we usually reserve the word
■
polynomial for expressions formed by more three terms.
The degree of a polynomial is the largest sum of the terms’ exponents.
■
Examples
The degree of the trinomial x2 + 2x – 1 is 2, because the x2 term has the highest exponent of 2.
■
The degree of the binomial x + 2 is 1, because the x term has the highest exponent of 1.
■
The degree of the binomial –3x4y2 + 2xy is 6, because the x4y2 term has the highest exponent sum of 6.
■
L IKE T ERMS
If two or more terms have exactly the same variable(s), and these variables are raised to exactly the same expo-
nents, they are said to be like terms. Like terms can be simplified when added and subtracted.
Examples
7x + 3x = 10x
6y2 – 4y2 = 2y2
3cd2 + 5c2d cannot be simplified. Since the exponent of 2 is on d in 3cd2 and on c in 5c2d, they are not like
terms.
The process of adding and subtracting like terms is called combining like terms. It is important to combine
like terms carefully, making sure that the variables are exactly the same.
Algebraic Expressions
An algebraic expression is a combination of monomials and operations. The difference between algebraic expres-
sions and algebraic equations is that algebraic expressions are evaluated at different given values for variables, while
algebraic equations are solved to determine the value of the variable that makes the equation a true statement.
There is very little difference between expressions and equations, because equations are nothing more than
two expressions set equal to each other. Their usage is subtly different.
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Example
A mobile phone company charges a $39.99 a month flat fee for the first 600 minutes of calls, with a
charge of $.55 for each minute thereafter.
Write an algebraic expression for the cost of a month’s mobile phone bill:
$39.99 + $.55x, where x represents the number of additional minutes used.
Write an equation for the cost (C) of a month’s mobile phone bill:
C = $39.99 + $.55x, where x represents the number of additional minutes used.
In the above example, you might use the expression $39.99 + $.55x to determine the cost if you are given
the value of x by substituting the value for x. You could also use the equation C = $39.99 + $.55x in the same way,
but you can also use the equation to determine the value of x if you were given the cost.
S IMPLIFYING E VALUATING A LGEBRAIC E XPRESSIONS
AND
We can use the mobile phone company example above to illustrate how to simplify algebraic expressions. Alge-
braic expressions are evaluated by a two-step process; substituting the given value(s) into the expression, and then
simplifying the expression by following the order of operations (PEMDAS).
Example
Using the cost expression $39.99 + $.55x, determine the total cost of a month’s mobile phone bill if
the owner made 700 minutes of calls.
Let x represent the number of minutes over 600 used, so in order to find out the difference, subtract
700 – 600; x = 100 minutes over 600 used.
Substitution: Replace x with its value, using parentheses around the value.
$39.99 + $.55x
$39.99 + $.55(100)
Evaluation: PEMDAS tells us to evaluate Parentheses and Exponents first. There is no operation to perform
in the parentheses, and there are no exponents, so the next step is to multiply, and then add.
$39.99 + $.55(100)
$39.99 + $55 = $94.99
The cost of the mobile phone bill for the month is $94.99.
You can evaluate algebraic expressions that contain any number of variables, as long as you are given all of
the values for all of the variables.
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S imple Rules for Working with Linear Equations
A linear equation is an equation whose variables’ highest exponent is 1. It is also called a first-degree equation.
An equation is solved by finding the value of an unknown variable.
1. The equal sign separates an equation into two sides.
2. Whenever an operation is performed on one side, the same operation must be performed on the other
side.
3. The first goal is to get all of the variable terms on one side and all of the numbers (called constants) on
the other side. This is accomplished by undoing the operations that are attaching numbers to the variable,
thereby isolating the variable. The operations are always done in reverse “PEMDAS” order: start by
adding/subtracting, then multiply/divide.
4. The final step often will be to divide each side by the coefficient, the number in front of the variable, leav-
ing the variable alone and equal to a number.
Example
5m + 8 = 48
–8 = –8
5m 40
5=5
m= 8
Undo the addition of 8 by subtracting 8 from both sides of the equation. Then undo the multiplication by
5 by dividing by 5 on both sides of the equation. The variable, m, is now isolated on the left side of the equation,
and its value is 8.
Checking Solutions to Equations
To check an equation, substitute the value of the variable into the original equation.
Example
To check the solution of the previous equation, substitute the number 8 for the variable m in
5m + 8 = 48.
5(8) + 8 = 48
40 + 8 = 48
48 = 48
Because this statement is true, the answer m = 8 must be correct.
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I SOLATING VARIABLES U SING F RACTIONS
Working with equations that contain fractions is almost exactly the same as working with equations that do not
contain variables, except for the final step. The final step when an equation has no fractions is to divide each side
by the coefficient. When the coefficient of the variable is a fraction, you will instead multiply both sides by the recip-
rocal of the coefficient. Technically, you could still divide both sides by the coefficient, but that involves division
of fractions which can be trickier.
Example
2 1
3m + 2 = 12
1 1
–2 =– 2
2 1
3m = 11 2
32 13
2 • 3m = 11 2 • 2
32 23 3
2 • 3m = 2 •2
69
m= 4
1 1
Undo the addition of 2 by subtracting 2 from both sides of the equation. Multiply both sides by the recip-
1
rocal of the coefficient. Convert the 11 2 to an improper fraction to facilitate multiplication. The variable m is now
isolated on the left side of the equation, and its value is 649 .
Equations with More than One Variable
Equations can have more than one variable. Each variable represents a different value, although it is possible that
the variables have the same value.
Remember that like terms have the same variable and exponent. All of the rules for working with variables
apply in equations that contain more than one variable, but you must remember not to combine terms that are
not alike.
Equations with more than one variable cannot be “solved,”because if there is more than one variable in an equa-
tion there is usually an infinite number of values for the variables that would make the equation true. Instead, we are
often required to “solve for a variable,” which instead means to isolate that variable on one side of the equation.
Example
Solve for y in the equation 2x + 3y = 5.
There are an infinite number of values for x and y that that satisfy the equation. Instead, we are
asked to isolate y on one side of the equation.
2x + 3y = 5
– 2x = – 2x
3y – 2x + 5
=
3 3
–2x + 5
y= 3
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C ross Multiplying
Since algebra uses percents and proportions, it is necessary to learn how to cross multiply. You can solve an equa-
tion that sets one fraction equal to another by cross multiplication. Cross multiplication involves setting the cross
products of opposite pairs of terms equal to each other.
Example
x 70
10 = 100
100x = 700
100x 700
100 = 100
x=7
Algebraic Fractions
Working with algebraic fractions is very similar to working with fractions in arithmetic. The difference is that alge-
braic fractions contain algebraic expressions in the numerator and/or denominator.
Example
A hotel currently has only one-fifth of their rooms available. If x represents the total number of
rooms in the hotel, find an expression for the number of rooms that will be available if another
tenth of the total rooms are reserved.
x 1
Since x represents the total number of rooms, 5 (or 5 x) represents the number of available rooms.
x
One tenth of the total rooms in the hotel would be represented by the fraction 10 . To find the new
x x
number of available rooms, find the difference: 5 – 10 .
x x
Write 5 – 10 as a single fraction.
Just like in arithmetic, the first step is to find the LCD of 5 and 10, which is 10. Then change each
fraction into an equivalent fraction that has 10 as a denominator.
x x x(2) x
– = 5(2) – 10
5 10
2x x
= 10 – 10
x
= 10
x
Therefore, rooms will be available after another tenth of the rooms are reserved.
10
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R eciprocal Rules
1
There are special rules for the sum and difference of reciprocals. The reciprocal of 3 is and the reciprocal
3
1
of x is x .
1 1 y x y+x
If x and y are not 0, then x + y = xy + xy = xy .
■
1 1 y x y–x
If x and y are not 0, then x – y = xy – xy = xy .
■
Translating Words into Numbers
The most important skill needed for word problems is being able to translate words into mathematical operations.
The following will be helpful in achieving this goal by providing common examples of English phrases and their
mathematical equivalents.
Phrases meaning addition: increased by; sum of; more than; exceeds by.
Examples
A number increased by five: x + 5.
The sum of two numbers: x + y.
Ten more than a number: x + 10.
Phrases meaning subtraction: decreased by; difference of; less than; diminished by.
Examples
10 less than a number: x – 10.
The difference of two numbers: x – y.
Phrases meaning multiplication: times; times the sum/difference; product; of.
Examples
Three times a number: 3x.
Twenty percent of 50: 20% 50.
Five times the sum of a number and three: 5(x + 3).
Phrases meaning “equals”: is; result is.
Examples
15 is 14 plus 1: 15 = 14 + 1.
10 more than 2 times a number is 15: 2x + 10 = 15.
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A ssigning Variables in Word Problems
It may be necessary to create and assign variables in a word problem. To do this, first identify any knowns and
unknowns. The known may not be a specific numerical value, but the problem should indicate something about
its value. Then let x represent the unknown you know the least about.
Examples
Max has worked for three more years than Ricky.
Unknown: Ricky’s work experience = x
Known: Max’s experience is three more years = x + 3
Heidi made twice as many sales as Rebecca.
Unknown: number of sales Rebecca made = x
Known: number of sales Heidi made is twice Rebecca’s amount = 2x
There are six less than four times the number of pens than pencils.
Unknown: the number of pencils = x
Known: the number of pens = 4x – 6
Todd has assembled five more than three times the number of cabinets that Andrew has.
Unknown: the number of cabinets Andrew has assembled = x
Known: the number of cabinets Todd has assembled is five more than 3 times the number Andrew
has assembled = 3x + 5
Percentage Problems
To solve percentage problems, determine what information has been given in the problem and fill this informa-
tion into the following template:
____ is ____% of ____
Then translate this information into a one-step equation and solve. In translating, remember that is trans-
lates to “=” and of translates to “ ”. Use a variable to represent the unknown quantity.
Examples
A) Finding a percentage of a given number:
In a new housing development there will be 50 houses; 40% of the houses must be completed in the
first stage. How many houses are in the first stage?
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1. Translate.
____ is 40% of 50.
x is .40 50.
2. Solve.
x = .40 50
x = 20
20 is 40% of 50. There are 20 houses in the first stage.
B) Finding a number when a percentage is given:
40% of the cars on the lot have been sold. If 24 were sold, how many total cars are there on the lot?
1. Translate.
24 is 40% of ____.
24 = .40 x.
2. Solve.
24 .40x
.40 = .40
60 = x
24 is 40% of 60. There were 60 total cars on the lot.
C) Finding what percentage one number is of another:
Matt has 75 employees. He is going to give 15 of them raises. What percentage of the employees will
receive raises?
1. Translate.
15 is ____% of 75.
15 = x 75.
2. Solve.
15 75x
= 75
75
.20 = x
20% = x
15 is 20% of 75. Therefore, 20% of the employees will receive raises.
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P roblems Involving Ratio
A ratio is a comparison of two quantities measured in the same units. It is symbolized by the use of a colon—x:y.
Ratios can also be expressed as fractions ( x ) or using words (x to y).
y
Ratio problems are solved using the concept of multiples.
Example
A bag contains 60 screws and nails. The ratio of the number of screws to nails is 7:8. How many of
each kind are there in the bag?
From the problem, it is known that 7 and 8 share a multiple and that the sum of their product is 60.
Whenever you see the word ratio in a problem, place an “x” next to each of the numbers in the ratio,
and those are your unknowns.
Let 7x = the number of screws.
Let 8x = the number of nails.
Write and solve the following equation:
7x + 8x = 60
15x 60
= 15
15
x=4
Therefore, there are (7)(4) = 28 screws and (8)(4) = 32 nails.
28 7
Check: 28 + 32 = 60 screws, 32 = 8 .
Problems Involving Variation
Variation is a term referring to a constant ratio in the change of a quantity.
Two quantities are said to vary directly if their ratios are constant. Both variables change in an equal
■
direction. In other words, two quantities vary directly if an increase in one causes an increase in the other.
This is also true if a decrease in one causes a decrease in the other.
Example
If it takes 300 new employees a total of 58.5 hours to train, how many hours of training will it take
for 800 employees?
Since each employee needs about the same amount of training, you know that they vary directly.
Therefore, you can set the problem up the following way:
300 800
employees
→ =
hours 58.5 x
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Cross-multiply to solve:
(800)(58.5)= 300x
300x
46,800
= 300
300
156 = x
Therefore, it would take 156 hours to train 800 employees.
Two quantities are said to vary inversely if their products are constant. The variables change in opposite
■
directions. This means that as one quantity increases, the other decreases, or as one decreases, the other
increases.
Example
If two people plant a field in six days, how many days will it take six people to plant the same field?
(Assume each person is working at the same rate.)
As the number of people planting increases, the days needed to plant decreases. Therefore, the
relationship between the number of people and days varies inversely. Because the field remains
constant, the two products can be set equal to each other.
2 people 6 days = 6 people x days
2 6 = 6x
12 6x
6=6
2=x
Thus, it would take 6 people 2 days to plant the same field.
Rate Problems
In general, there are three different types of rate problems likely to be encountered in the workplace: cost per unit,
movement, and work-output. Rate is defined as a comparison of two quantities with different units of measure.
x units
Rate = y units
Examples
dollars cost miles
hour , pound , hour
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C OST P ER U NIT
Some problems will require the calculation of unit cost.
Example
If 100 square feet cost $1,000, how much does 1 square foot cost?
$1,000
Total cost
= 100 ft2
# of square feet
= $10 per square foot
M OVEMENT
In working with movement problems, it is important to use the following formula:
(Rate)(Time) = Distance
Example
1
A courier traveling at 15 mph traveled from his base to a company in 4 of an hour less than it took
when the courier traveled 12 mph. How far away was his drop off?
First, write what is known and unknown.
Unknown: time for courier traveling 12 mph = x.
1
Known: time for courier traveling 15 mph = x – 4 .
Then, use the formula (Rate)(Time) = Distance to find expressions for the distance traveled at each
rate:
12 mph for x hours = a distance of 12x miles.
1 15
15 miles per hour for x – 4 hours = a distance of 15x – 4 miles.
The distance traveled is the same, therefore, make the two expressions equal to each other:
12x = 15x – 3.75
–15x = –15x
–3x –3.75
–3 = –3
x = 1.25
Be careful, 1.25 is not the distance; it is the time. Now you must plug the time into the formula
(Rate)(Time) = Distance. Either rate can be used.
12x = distance
12(1.25) = distance
15 miles = distance
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W ORK -O UTPUT
Work-output problems are word problems that deal with the rate of work. The following formula can be used on
these problems:
(Rate of Work)(Time Worked) = Job or Part of Job Completed
Example
Danette can wash and wax 2 cars in 6 hours, and Judy can wash and wax the same two cars in 4
hours. If Danette and Judy work together, how long will it take to wash and wax one car?
Since Danette can wash and wax 2 cars in 6 hours, her rate of work is 62hcaurs , or one car every three
rs
o
hours. Judy’s rate of work is therefore, 42hcaurs , or one car every two hours. In this problem, making a
rs
o
chart will help:
Rate Time = Part of job completed
1 1
Danette x = 3x
3
1 1
Judy x = 2x
2
Since they are both working on only one car, you can set the equation equal to one:
Danette’s part + Judy’s part = 1 car:
1 1
3x + 2x = 1
Solve by using 6 as the LCD for 3 and 2 and clear the fractions by multiplying by the LCD:
1 1
6( 3 x) + 6( 2 x) = 6(1)
2x + 3x = 6
5x 6
=
5 5
x = 11
5
Thus, it will take Judy and Danette 1 1 hours to wash and wax one car.
5
Patterns and Functions
The ability to detect patterns in numbers is a very important mathematical skill. Patterns exist everywhere in nature,
business, and finance.
When you are asked to find a pattern in a series of numbers, look to see if there is some common number
you can add, subtract, multiply, or divide each number in the pattern by to give you the next number in the series.
For example, in the sequence 5, 8, 11, 14 . . . you can add three to each number in the sequence to get the
next number in the sequence. The next number in the sequence is 17.
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Example
3
What is the next number in the sequence 4 , 3, 12, 48?
Each number in the sequence can be multiplied by the number 4 to get the next number in the
3
sequence: 4 4 = 3, 3 4 = 12, 12 4 = 48, so the next number in the sequence is 48 4 = 192.
Sometimes it is not that simple. You may need to look for a combination of multiplying and adding, divid-
ing and subtracting, or some combination of other operations.
Example
What is the next number in the sequence 0, 1, 2, 5, 26?
Keep trying various operations until you find one that works. In this case, the correct procedure is
to square the term and add 1: 02 + 1 = 1, 12 + 1 = 2, 22 + 1 = 5, 52 + 1 = 26, so the next number in
the sequence is 262 + 1 = 677.
P ROPERTIES F UNCTIONS
OF
A function is a relationship between two variables x and y where for each value of x, there is one and only one
value of y. Functions can be represented in four ways:
a table or chart
■
an equation
■
a word problem
■
a graph
■
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For example, the following four representations are equivalent to the same function:
Equation
Word Problem
Javier has one more than two times the
y = 2x + 1
number of books Susanna has.
Graph
y
5 Table
y
4 x
3 –5
–3
–3
2 –2
1 –1
–1
x 1
0
1
–5 2 3 4 5
–4 –3 –2 –1
–1 3
1
–2 5
2
–3
–4
–5
Helpful hints for determining if a relation is a function:
If you can isolate y in terms of x using only one equation, it is a function.
■
If the equation contains y2, it will not be a function.
■
If you can draw a vertical line anywhere on a graph such that it touches the graph in more than one place,
■
it is not a function.
If there is a value for x that has more than one y-value assigned to it, it is not a function.
■
133
- – THEA MATH REVIEW –
y y
x
y x
x
5 3
–2
2 –2
5
6 3
–1
–1 –1
3
7 3
0
0 0
2
8 3
1
1 1
6
9 3
2
4 2
5
In this table, every
In this table, every
In this table, the
x-value
x-value
x-value of 5 has two
{–2, –1, 0, 1, 2, 3}
{–2, –1, 0, 1, 2, 3}
corresponding y-values,
has one corresponding
has one corresponding
2 and 4. Therefore,
y-value, even though
y-value. This is
it is not a function.
that value is 3
a function.
in every case.
This is a function.
y
y
5
5
4
4
3
3
2
2
1
1
x
x
1
–5 2 3 4 5
1 –4 –3 –2 –1
–5 2 3 4 5
–4 –3 –2 –1
–1 –1
–2 –2
–3 –3
–4 –4
–5 –5
In this graph, there is no
In this graph, there
vertical line that can be drawn
is at least one vertical
that intersects the graph
line that can be drawn
in more than one place.
(the dotted line) that intersects
This is a function.
the graph in more than one
place. This is not a function.
134
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