���TEMPERATURE The equation to convert Celsius temperature C to Kelvin temperature K is shown. �

� a. State the independent and dependent variables.Explain. b. Determine the C- and K-intercepts and describe what the intercepts mean in this situation.

62/87,21���a. The Kelvin temperature is dependent on the Celsius temperature. So, the Celsius temperature is the independent variable and the Kelvin temperature is the dependent variable. � b. The C-intercept is (±273, 0) and it means that a Celsius temperature of ±273 degrees is equal to a Kelvin temperature of 0 degrees. The K-intercept is (0, 273) and it means that a Celsius temperature of 0 degrees is equal to a Kelvin temperature of 273 degrees.

Graph each equation.���y = x + 2

62/87,21���To graph the equation, find the x- and y-intercepts. Plot these two points. Then draw a line through them.To find the x-intercept, let y = 0. �

� To find the y-intercept, let x = 0. �

� So, the x-intercept is ±2 and the y-intercept is 2. �

���y = 4x

62/87,21���The slope of y = 4x is 4. Graph (0, 0). From there, move up 4 units and right 1 unit to find another point. Then draw aline containing the points. �

���x + 2y = ±1

62/87,21���To graph the equation, find the x- and y-intercepts. Plot these two points. Then draw a line through them.To find the x-intercept, let y = 0. �

� To find the y-intercept, let x = 0. �

So, the x-intercept is ±1 and the y-intercept is .

�

���±3x = 5 ± y

62/87,21���To graph the equation, find the x- and y-intercepts. Plot these two points. Then draw a line through them.To find the x-intercept, let y = 0. �

� To find the y-intercept, let x = 0. �

�

So, the x-intercept is �DQG�WKH�y-intercept is 5.

�

�

Solve each equation by graphing.���4x + 2 = 0

62/87,21���The related function is y = 4x + 2. 7R�JUDSK�WKH�IXQFWLRQ��PDNH�D�WDEOH���

The graph intersects the x-axis at . So, the solution is .

x f (x ) = 4x + 2 f (x �� (x, f(x ) ) ±4 f (±4) = 4(±4) + 2 ±14 (±4, ±14) ±2 f (±2) = 4(±2) + 2 ±6 (±2, ±6) ±0.5 f(±0.5) = 4(±0.5) + 2 0 (±0.5, 0)

0 f (0) = 4(0) + 2 2 (0, 2) 2 f (2) = 4(2) + 2 10 (2, 10) 4 f (4) = 4(4) + 2 18 (4, 18)

���0 = 6 ± 3x

62/87,21���The related function is y = í3x + 6. 7R�JUDSK�WKH�IXQFWLRQ��PDNH�D�WDEOH���

�

The graph intersects the x-axis at 2. So, the solution is 2.

x f (x ) = ±3x + 6 f (x �� (x, f(x ) ) ±4 f (±4) = ±3(±4) + 6 18 (±4, 18) ±2 f (±2) = ±3(±2) + 6 12 (±2, 12) 0 f (0) = ±3(0) + 6 6 (0, 6) 2 f (2) = ±3(2) + 6 0 (2, 0) 3 f (3) = ±3(3) + 6 ±3 (3, ±3) 4 f (4) = ±3(4) + 6 ±6 (4, ±6)

���5x + 2 = ±3

62/87,21���The related function is y = 5x + 5. 7R�JUDSK�WKH�IXQFWLRQ��PDNH�D�WDEOH���

�

The graph intersects the x-axis at ±1. So, the solution is ±1.

x f (x ) = 5x + 5 f (x �� (x, f(x ) ) ±4 f (±4) = 5(±4) + 5 ±15 (±4, ±15) ±2 f (±2) = 5(±2) + 5 ±5 (±2, ±5) ±1 f (±1) = 5(±1) + 5 0 (±1, 0) 0 f (0) = 5(0) + 5 5 (0, 5) 2 f (2) = 5(2) + 5 15 (2, 15) 4 f (4) = 5(4) + 5 25 (4, 25)

���12x = 4x + 16

62/87,21���The related function is y = í8x��������7R�JUDSK�WKH�IXQFWLRQ��PDNH�D�WDEOH���

�

The graph intersects the x-axis at 2. So, the solution is 2.

x f (x ) = ±8x + 16 f (x �� (x, f(x ) ) ±4 f (±4) = ±8(±4) + 16 48 (±4, 48) ±2 f (±2) = ±8(±2) + 16 32 (±2, 32) 0 f (0) = ±8(0) + 16 16 (0, 16) 2 f (2) = ±8(2) + 16 0 (2, 0) 3 f (3) = ±8(3) + 16 ±8 (3, ±8) 4 f (4) = ±8(4) + 16 ±16 (4, ±16)

Find the slope of the line that passes through each pair of points.����(5, 8), (±3, 7)

62/87,21���

So, the slope is .

����(5, ±2), (3, ±2)

62/87,21���

� So, the slope is 0.

����(±4, 7), (8, ±1)

62/87,21���

So, the slope is .

����(6, ±3), (6, 4)

62/87,21���

� So, the slope is undefined.

����MULTIPLE CHOICE Which is the slope of the linear function shown in the graph?�

A

B �

C

D

62/87,21���The line passes through the points (±2, 3) and (3, 1).�

The slope is , so the correct choice is B.

Suppose y varies directly as x. Write a direct variation equation that relates x and y . Then solve.����If y = 6 when x = 9, find x when y = 12.

62/87,21���

� So, the direct variation equation is ����Substitute 12 for y and find x.´��

� So, x = 18 when y = 12.

����When y = ±8, x = 8. What is x when y = ±6?

62/87,21���

� So, the direct variation equation is y = ±x���Substitute ±6 for y and find x.´��

� So, x = 6 when y = ±6.

����If y = ±5 when x = ±2, what is y when x = 14?

62/87,21���

� So, the direct variation equation is ���Substitute 14 for x and find y .�

� So, y = 35 when x = 14.

����If y = 2 when x = ±12, find y when x = ±4.

62/87,21���

� So, the direct variation equation is ����Substitute ±4 for x and find y .´��

�

So, y = �ZKHQ�x = ±4.

����BIOLOGY The number of pints of blood in a human body varies directly with the person¶s weight. A person who weighs 120 pounds has about 8.4 pints of blood in his or her body. a. Write and graph an equation relating weight and amount of blood in a person¶s body. b. Predict the weight of a person whose body holds 12 pints of blood.

62/87,21���a. To write a direct variation equation, find the constant of variation k . Let x = 120 and y = 8.4.�

� So, the direct variation equation is y = 0.07x.

b. Using the direct variation equation from part a, let y = 12. �

� So, a person who holds 12 pints of blood would weigh about 171 pounds.

Find the next three terms of each arithmetic sequence.����0, ±15, ±30, ±45, ±60, «

62/87,21���Find the common difference by subtracting two consecutive terms. � ±15 ± 0 = ±���� � The common difference between terms is ±�����6R��WR�ILQG�WKH�QH[W�WHUP��VXEWUDFW����IURP�WKH�ODVW�WHUP��7R�ILQG�WKH�next term, subtract 15 from the resulting number, and so on. � ±60 ± 15 = ±75 ±75 ± 15 = ±90 ±90 ± 15 = ±105 � So, the next three terms of this arithmetic sequence are ±75, ±90, ±105.

����5, 8, 11, 14, «

62/87,21���Find the common difference by subtracting two consecutive terms. � 8 ± �� ���� � The common difference between terms is 3. So, to find the next term, add 3 from the last term. To find the next term, add 3 from the resulting number, and so on. � 14 + 3 = 17 17 + 3 = 20 20 + 3 = 23 � So, the next three terms of this arithmetic sequence are 17, 20, 23.

Determine whether each sequence is an arithmetic sequence. If it is, state the common difference.����±40, ±32, ±24, ±16, «

62/87,21���±40, ±32, ±24, ±16, « � An arithmetic sequence is a numerical pattern that increases or decreases at a constant rate called the common difference. To find the common difference, subtract two consecutive numbers in the sequence. � ±32 ± (±40) = 8 ±24 ± (±32) = 8 ±16 ± (±24) = 7 � 7KH�GLIIHUHQFH�EHWZHHQ�WHUPV�LV�FRQVWDQW��VR�WKH�VHTXHQFH�LV�DQ�DULWKPHWLF�VHTXHQFH�� The common difference is 4. �

����0.75, 1.5, 3, 6, 12, «

62/87,21���An arithmetic sequence is a numerical pattern that increases or decreases at a constant rate called the common difference. To find the common difference, subtract two consecutive numbers in the sequence. � 1.5 ± 0.75 = 0.75 3 ± 1.5 = 1.5 6 ± 3 = 3 12 ± 6 = 6 � The difference between terms is not constant, so the sequence is not an an arithmetic sequence and does not have a common difference�� �

����5, 17, 29, 41, «

62/87,21���An arithmetic sequence is a numerical pattern that increases or decreases at a constant rate called the common difference. To find the common difference, subtract two consecutive numbers in the sequence. � 17 ± 5 = 12 29 ± 17 = 12 41 ± 29 = 12 � 7KH�GLIIHUHQFH�EHWZHHQ�WHUPV�LV�FRQVWDQW��VR�WKH�VHTXHQFH�LV�DQ�DULWKPHWLF�VHTXHQFH�� The common difference is 4. �

����MULTIPLE CHOICE In each figure, only one side of each regular pentagon is shared with another pentagon. Each side of each pentagon is 1 centimeter. If the pattern continues, what is the perimeter of a figure that has 6 pentagons?

F 15 cm H 20 cm G 25 cm J 30 cm

62/87,21���

� The perimeter increases by 3 as the number of polygons increases by 1. Continue the sequence to find the perimeter of a figure that has 6 pentagons. �

� So, a figure that has 6 pentagons has a perimeter of 20 centimeters. The correct choice is H.

Number of Pentagons

Perimeter

1 5 2 8 3 11

Number of Pentagons

Perimeter

4 14 5 17 6 20

eSolutions Manual - Powered by Cognero Page 1

Practice Test - Chapter 3

���TEMPERATURE The equation to convert Celsius temperature C to Kelvin temperature K is shown. �

� a. State the independent and dependent variables.Explain. b. Determine the C- and K-intercepts and describe what the intercepts mean in this situation.

62/87,21���a. The Kelvin temperature is dependent on the Celsius temperature. So, the Celsius temperature is the independent variable and the Kelvin temperature is the dependent variable. � b. The C-intercept is (±273, 0) and it means that a Celsius temperature of ±273 degrees is equal to a Kelvin temperature of 0 degrees. The K-intercept is (0, 273) and it means that a Celsius temperature of 0 degrees is equal to a Kelvin temperature of 273 degrees.

Graph each equation.���y = x + 2

62/87,21���To graph the equation, find the x- and y-intercepts. Plot these two points. Then draw a line through them.To find the x-intercept, let y = 0. �

� To find the y-intercept, let x = 0. �

� So, the x-intercept is ±2 and the y-intercept is 2. �

���y = 4x

62/87,21���The slope of y = 4x is 4. Graph (0, 0). From there, move up 4 units and right 1 unit to find another point. Then draw aline containing the points. �

���x + 2y = ±1

62/87,21���To graph the equation, find the x- and y-intercepts. Plot these two points. Then draw a line through them.To find the x-intercept, let y = 0. �

� To find the y-intercept, let x = 0. �

So, the x-intercept is ±1 and the y-intercept is .

�

���±3x = 5 ± y

62/87,21���To graph the equation, find the x- and y-intercepts. Plot these two points. Then draw a line through them.To find the x-intercept, let y = 0. �

� To find the y-intercept, let x = 0. �

�

So, the x-intercept is �DQG�WKH�y-intercept is 5.

�

�

Solve each equation by graphing.���4x + 2 = 0

62/87,21���The related function is y = 4x + 2. 7R�JUDSK�WKH�IXQFWLRQ��PDNH�D�WDEOH���

The graph intersects the x-axis at . So, the solution is .

x f (x ) = 4x + 2 f (x �� (x, f(x ) ) ±4 f (±4) = 4(±4) + 2 ±14 (±4, ±14) ±2 f (±2) = 4(±2) + 2 ±6 (±2, ±6) ±0.5 f(±0.5) = 4(±0.5) + 2 0 (±0.5, 0)

0 f (0) = 4(0) + 2 2 (0, 2) 2 f (2) = 4(2) + 2 10 (2, 10) 4 f (4) = 4(4) + 2 18 (4, 18)

���0 = 6 ± 3x

62/87,21���The related function is y = í3x + 6. 7R�JUDSK�WKH�IXQFWLRQ��PDNH�D�WDEOH���

�

The graph intersects the x-axis at 2. So, the solution is 2.

x f (x ) = ±3x + 6 f (x �� (x, f(x ) ) ±4 f (±4) = ±3(±4) + 6 18 (±4, 18) ±2 f (±2) = ±3(±2) + 6 12 (±2, 12) 0 f (0) = ±3(0) + 6 6 (0, 6) 2 f (2) = ±3(2) + 6 0 (2, 0) 3 f (3) = ±3(3) + 6 ±3 (3, ±3) 4 f (4) = ±3(4) + 6 ±6 (4, ±6)

���5x + 2 = ±3

62/87,21���The related function is y = 5x + 5. 7R�JUDSK�WKH�IXQFWLRQ��PDNH�D�WDEOH���

�

The graph intersects the x-axis at ±1. So, the solution is ±1.

x f (x ) = 5x + 5 f (x �� (x, f(x ) ) ±4 f (±4) = 5(±4) + 5 ±15 (±4, ±15) ±2 f (±2) = 5(±2) + 5 ±5 (±2, ±5) ±1 f (±1) = 5(±1) + 5 0 (±1, 0) 0 f (0) = 5(0) + 5 5 (0, 5) 2 f (2) = 5(2) + 5 15 (2, 15) 4 f (4) = 5(4) + 5 25 (4, 25)

���12x = 4x + 16

62/87,21���The related function is y = í8x��������7R�JUDSK�WKH�IXQFWLRQ��PDNH�D�WDEOH���

�

The graph intersects the x-axis at 2. So, the solution is 2.

x f (x ) = ±8x + 16 f (x �� (x, f(x ) ) ±4 f (±4) = ±8(±4) + 16 48 (±4, 48) ±2 f (±2) = ±8(±2) + 16 32 (±2, 32) 0 f (0) = ±8(0) + 16 16 (0, 16) 2 f (2) = ±8(2) + 16 0 (2, 0) 3 f (3) = ±8(3) + 16 ±8 (3, ±8) 4 f (4) = ±8(4) + 16 ±16 (4, ±16)

Find the slope of the line that passes through each pair of points.����(5, 8), (±3, 7)

62/87,21���

So, the slope is .

����(5, ±2), (3, ±2)

62/87,21���

� So, the slope is 0.

����(±4, 7), (8, ±1)

62/87,21���

So, the slope is .

����(6, ±3), (6, 4)

62/87,21���

� So, the slope is undefined.

����MULTIPLE CHOICE Which is the slope of the linear function shown in the graph?�

A

B �

C

D

62/87,21���The line passes through the points (±2, 3) and (3, 1).�

The slope is , so the correct choice is B.

Suppose y varies directly as x. Write a direct variation equation that relates x and y . Then solve.����If y = 6 when x = 9, find x when y = 12.

62/87,21���

� So, the direct variation equation is ����Substitute 12 for y and find x.´��

� So, x = 18 when y = 12.

����When y = ±8, x = 8. What is x when y = ±6?

62/87,21���

� So, the direct variation equation is y = ±x���Substitute ±6 for y and find x.´��

� So, x = 6 when y = ±6.

����If y = ±5 when x = ±2, what is y when x = 14?

62/87,21���

� So, the direct variation equation is ���Substitute 14 for x and find y .�

� So, y = 35 when x = 14.

����If y = 2 when x = ±12, find y when x = ±4.

62/87,21���

� So, the direct variation equation is ����Substitute ±4 for x and find y .´��

�

So, y = �ZKHQ�x = ±4.

����BIOLOGY The number of pints of blood in a human body varies directly with the person¶s weight. A person who weighs 120 pounds has about 8.4 pints of blood in his or her body. a. Write and graph an equation relating weight and amount of blood in a person¶s body. b. Predict the weight of a person whose body holds 12 pints of blood.

62/87,21���a. To write a direct variation equation, find the constant of variation k . Let x = 120 and y = 8.4.�

� So, the direct variation equation is y = 0.07x.

b. Using the direct variation equation from part a, let y = 12. �

� So, a person who holds 12 pints of blood would weigh about 171 pounds.

Find the next three terms of each arithmetic sequence.����0, ±15, ±30, ±45, ±60, «

62/87,21���Find the common difference by subtracting two consecutive terms. � ±15 ± 0 = ±���� � The common difference between terms is ±�����6R��WR�ILQG�WKH�QH[W�WHUP��VXEWUDFW����IURP�WKH�ODVW�WHUP��7R�ILQG�WKH�next term, subtract 15 from the resulting number, and so on. � ±60 ± 15 = ±75 ±75 ± 15 = ±90 ±90 ± 15 = ±105 � So, the next three terms of this arithmetic sequence are ±75, ±90, ±105.

����5, 8, 11, 14, «

62/87,21���Find the common difference by subtracting two consecutive terms. � 8 ± �� ���� � The common difference between terms is 3. So, to find the next term, add 3 from the last term. To find the next term, add 3 from the resulting number, and so on. � 14 + 3 = 17 17 + 3 = 20 20 + 3 = 23 � So, the next three terms of this arithmetic sequence are 17, 20, 23.

Determine whether each sequence is an arithmetic sequence. If it is, state the common difference.����±40, ±32, ±24, ±16, «

62/87,21���±40, ±32, ±24, ±16, « � An arithmetic sequence is a numerical pattern that increases or decreases at a constant rate called the common difference. To find the common difference, subtract two consecutive numbers in the sequence. � ±32 ± (±40) = 8 ±24 ± (±32) = 8 ±16 ± (±24) = 7 � 7KH�GLIIHUHQFH�EHWZHHQ�WHUPV�LV�FRQVWDQW��VR�WKH�VHTXHQFH�LV�DQ�DULWKPHWLF�VHTXHQFH�� The common difference is 4. �

����0.75, 1.5, 3, 6, 12, «

62/87,21���An arithmetic sequence is a numerical pattern that increases or decreases at a constant rate called the common difference. To find the common difference, subtract two consecutive numbers in the sequence. � 1.5 ± 0.75 = 0.75 3 ± 1.5 = 1.5 6 ± 3 = 3 12 ± 6 = 6 � The difference between terms is not constant, so the sequence is not an an arithmetic sequence and does not have a common difference�� �

����5, 17, 29, 41, «

62/87,21���An arithmetic sequence is a numerical pattern that increases or decreases at a constant rate called the common difference. To find the common difference, subtract two consecutive numbers in the sequence. � 17 ± 5 = 12 29 ± 17 = 12 41 ± 29 = 12 � 7KH�GLIIHUHQFH�EHWZHHQ�WHUPV�LV�FRQVWDQW��VR�WKH�VHTXHQFH�LV�DQ�DULWKPHWLF�VHTXHQFH�� The common difference is 4. �

����MULTIPLE CHOICE In each figure, only one side of each regular pentagon is shared with another pentagon. Each side of each pentagon is 1 centimeter. If the pattern continues, what is the perimeter of a figure that has 6 pentagons?

F 15 cm H 20 cm G 25 cm J 30 cm

62/87,21���

� The perimeter increases by 3 as the number of polygons increases by 1. Continue the sequence to find the perimeter of a figure that has 6 pentagons. �

� So, a figure that has 6 pentagons has a perimeter of 20 centimeters. The correct choice is H.

Number of Pentagons

Perimeter

1 5 2 8 3 11

Number of Pentagons

Perimeter

4 14 5 17 6 20

eSolutions Manual - Powered by Cognero Page 2

Practice Test - Chapter 3

���TEMPERATURE The equation to convert Celsius temperature C to Kelvin temperature K is shown. �

� a. State the independent and dependent variables.Explain. b. Determine the C- and K-intercepts and describe what the intercepts mean in this situation.

62/87,21���a. The Kelvin temperature is dependent on the Celsius temperature. So, the Celsius temperature is the independent variable and the Kelvin temperature is the dependent variable. � b. The C-intercept is (±273, 0) and it means that a Celsius temperature of ±273 degrees is equal to a Kelvin temperature of 0 degrees. The K-intercept is (0, 273) and it means that a Celsius temperature of 0 degrees is equal to a Kelvin temperature of 273 degrees.

Graph each equation.���y = x + 2

62/87,21���To graph the equation, find the x- and y-intercepts. Plot these two points. Then draw a line through them.To find the x-intercept, let y = 0. �

� To find the y-intercept, let x = 0. �

� So, the x-intercept is ±2 and the y-intercept is 2. �

���y = 4x

62/87,21���The slope of y = 4x is 4. Graph (0, 0). From there, move up 4 units and right 1 unit to find another point. Then draw aline containing the points. �

���x + 2y = ±1

62/87,21���To graph the equation, find the x- and y-intercepts. Plot these two points. Then draw a line through them.To find the x-intercept, let y = 0. �

� To find the y-intercept, let x = 0. �

So, the x-intercept is ±1 and the y-intercept is .

�

���±3x = 5 ± y

62/87,21���To graph the equation, find the x- and y-intercepts. Plot these two points. Then draw a line through them.To find the x-intercept, let y = 0. �

� To find the y-intercept, let x = 0. �

�

So, the x-intercept is �DQG�WKH�y-intercept is 5.

�

�

Solve each equation by graphing.���4x + 2 = 0

62/87,21���The related function is y = 4x + 2. 7R�JUDSK�WKH�IXQFWLRQ��PDNH�D�WDEOH���

The graph intersects the x-axis at . So, the solution is .

x f (x ) = 4x + 2 f (x �� (x, f(x ) ) ±4 f (±4) = 4(±4) + 2 ±14 (±4, ±14) ±2 f (±2) = 4(±2) + 2 ±6 (±2, ±6) ±0.5 f(±0.5) = 4(±0.5) + 2 0 (±0.5, 0)

0 f (0) = 4(0) + 2 2 (0, 2) 2 f (2) = 4(2) + 2 10 (2, 10) 4 f (4) = 4(4) + 2 18 (4, 18)

���0 = 6 ± 3x

62/87,21���The related function is y = í3x + 6. 7R�JUDSK�WKH�IXQFWLRQ��PDNH�D�WDEOH���

�

The graph intersects the x-axis at 2. So, the solution is 2.

x f (x ) = ±3x + 6 f (x �� (x, f(x ) ) ±4 f (±4) = ±3(±4) + 6 18 (±4, 18) ±2 f (±2) = ±3(±2) + 6 12 (±2, 12) 0 f (0) = ±3(0) + 6 6 (0, 6) 2 f (2) = ±3(2) + 6 0 (2, 0) 3 f (3) = ±3(3) + 6 ±3 (3, ±3) 4 f (4) = ±3(4) + 6 ±6 (4, ±6)

���5x + 2 = ±3

62/87,21���The related function is y = 5x + 5. 7R�JUDSK�WKH�IXQFWLRQ��PDNH�D�WDEOH���

�

The graph intersects the x-axis at ±1. So, the solution is ±1.

x f (x ) = 5x + 5 f (x �� (x, f(x ) ) ±4 f (±4) = 5(±4) + 5 ±15 (±4, ±15) ±2 f (±2) = 5(±2) + 5 ±5 (±2, ±5) ±1 f (±1) = 5(±1) + 5 0 (±1, 0) 0 f (0) = 5(0) + 5 5 (0, 5) 2 f (2) = 5(2) + 5 15 (2, 15) 4 f (4) = 5(4) + 5 25 (4, 25)

���12x = 4x + 16

62/87,21���The related function is y = í8x��������7R�JUDSK�WKH�IXQFWLRQ��PDNH�D�WDEOH���

�

The graph intersects the x-axis at 2. So, the solution is 2.

x f (x ) = ±8x + 16 f (x �� (x, f(x ) ) ±4 f (±4) = ±8(±4) + 16 48 (±4, 48) ±2 f (±2) = ±8(±2) + 16 32 (±2, 32) 0 f (0) = ±8(0) + 16 16 (0, 16) 2 f (2) = ±8(2) + 16 0 (2, 0) 3 f (3) = ±8(3) + 16 ±8 (3, ±8) 4 f (4) = ±8(4) + 16 ±16 (4, ±16)

Find the slope of the line that passes through each pair of points.����(5, 8), (±3, 7)

62/87,21���

So, the slope is .

����(5, ±2), (3, ±2)

62/87,21���

� So, the slope is 0.

����(±4, 7), (8, ±1)

62/87,21���

So, the slope is .

����(6, ±3), (6, 4)

62/87,21���

� So, the slope is undefined.

����MULTIPLE CHOICE Which is the slope of the linear function shown in the graph?�

A

B �

C

D

62/87,21���The line passes through the points (±2, 3) and (3, 1).�

The slope is , so the correct choice is B.

Suppose y varies directly as x. Write a direct variation equation that relates x and y . Then solve.����If y = 6 when x = 9, find x when y = 12.

62/87,21���

� So, the direct variation equation is ����Substitute 12 for y and find x.´��

� So, x = 18 when y = 12.

����When y = ±8, x = 8. What is x when y = ±6?

62/87,21���

� So, the direct variation equation is y = ±x���Substitute ±6 for y and find x.´��

� So, x = 6 when y = ±6.

����If y = ±5 when x = ±2, what is y when x = 14?

62/87,21���

� So, the direct variation equation is ���Substitute 14 for x and find y .�

� So, y = 35 when x = 14.

����If y = 2 when x = ±12, find y when x = ±4.

62/87,21���

� So, the direct variation equation is ����Substitute ±4 for x and find y .´��

�

So, y = �ZKHQ�x = ±4.

����BIOLOGY The number of pints of blood in a human body varies directly with the person¶s weight. A person who weighs 120 pounds has about 8.4 pints of blood in his or her body. a. Write and graph an equation relating weight and amount of blood in a person¶s body. b. Predict the weight of a person whose body holds 12 pints of blood.

62/87,21���a. To write a direct variation equation, find the constant of variation k . Let x = 120 and y = 8.4.�

� So, the direct variation equation is y = 0.07x.

b. Using the direct variation equation from part a, let y = 12. �

� So, a person who holds 12 pints of blood would weigh about 171 pounds.

Find the next three terms of each arithmetic sequence.����0, ±15, ±30, ±45, ±60, «

62/87,21���Find the common difference by subtracting two consecutive terms. � ±15 ± 0 = ±���� � The common difference between terms is ±�����6R��WR�ILQG�WKH�QH[W�WHUP��VXEWUDFW����IURP�WKH�ODVW�WHUP��7R�ILQG�WKH�next term, subtract 15 from the resulting number, and so on. � ±60 ± 15 = ±75 ±75 ± 15 = ±90 ±90 ± 15 = ±105 � So, the next three terms of this arithmetic sequence are ±75, ±90, ±105.

����5, 8, 11, 14, «

62/87,21���Find the common difference by subtracting two consecutive terms. � 8 ± �� ���� � The common difference between terms is 3. So, to find the next term, add 3 from the last term. To find the next term, add 3 from the resulting number, and so on. � 14 + 3 = 17 17 + 3 = 20 20 + 3 = 23 � So, the next three terms of this arithmetic sequence are 17, 20, 23.

Determine whether each sequence is an arithmetic sequence. If it is, state the common difference.����±40, ±32, ±24, ±16, «

62/87,21���±40, ±32, ±24, ±16, « � An arithmetic sequence is a numerical pattern that increases or decreases at a constant rate called the common difference. To find the common difference, subtract two consecutive numbers in the sequence. � ±32 ± (±40) = 8 ±24 ± (±32) = 8 ±16 ± (±24) = 7 � 7KH�GLIIHUHQFH�EHWZHHQ�WHUPV�LV�FRQVWDQW��VR�WKH�VHTXHQFH�LV�DQ�DULWKPHWLF�VHTXHQFH�� The common difference is 4. �

����0.75, 1.5, 3, 6, 12, «

62/87,21���An arithmetic sequence is a numerical pattern that increases or decreases at a constant rate called the common difference. To find the common difference, subtract two consecutive numbers in the sequence. � 1.5 ± 0.75 = 0.75 3 ± 1.5 = 1.5 6 ± 3 = 3 12 ± 6 = 6 � The difference between terms is not constant, so the sequence is not an an arithmetic sequence and does not have a common difference�� �

����5, 17, 29, 41, «

62/87,21���An arithmetic sequence is a numerical pattern that increases or decreases at a constant rate called the common difference. To find the common difference, subtract two consecutive numbers in the sequence. � 17 ± 5 = 12 29 ± 17 = 12 41 ± 29 = 12 � 7KH�GLIIHUHQFH�EHWZHHQ�WHUPV�LV�FRQVWDQW��VR�WKH�VHTXHQFH�LV�DQ�DULWKPHWLF�VHTXHQFH�� The common difference is 4. �

����MULTIPLE CHOICE In each figure, only one side of each regular pentagon is shared with another pentagon. Each side of each pentagon is 1 centimeter. If the pattern continues, what is the perimeter of a figure that has 6 pentagons?

F 15 cm H 20 cm G 25 cm J 30 cm

62/87,21���

� The perimeter increases by 3 as the number of polygons increases by 1. Continue the sequence to find the perimeter of a figure that has 6 pentagons. �

� So, a figure that has 6 pentagons has a perimeter of 20 centimeters. The correct choice is H.

Number of Pentagons

Perimeter

1 5 2 8 3 11

Number of Pentagons

Perimeter

4 14 5 17 6 20

eSolutions Manual - Powered by Cognero Page 3

Practice Test - Chapter 3

���TEMPERATURE The equation to convert Celsius temperature C to Kelvin temperature K is shown. �

� a. State the independent and dependent variables.Explain. b. Determine the C- and K-intercepts and describe what the intercepts mean in this situation.

62/87,21���a. The Kelvin temperature is dependent on the Celsius temperature. So, the Celsius temperature is the independent variable and the Kelvin temperature is the dependent variable. � b. The C-intercept is (±273, 0) and it means that a Celsius temperature of ±273 degrees is equal to a Kelvin temperature of 0 degrees. The K-intercept is (0, 273) and it means that a Celsius temperature of 0 degrees is equal to a Kelvin temperature of 273 degrees.

Graph each equation.���y = x + 2

62/87,21���To graph the equation, find the x- and y-intercepts. Plot these two points. Then draw a line through them.To find the x-intercept, let y = 0. �

� To find the y-intercept, let x = 0. �

� So, the x-intercept is ±2 and the y-intercept is 2. �

���y = 4x

62/87,21���The slope of y = 4x is 4. Graph (0, 0). From there, move up 4 units and right 1 unit to find another point. Then draw aline containing the points. �

���x + 2y = ±1

62/87,21���To graph the equation, find the x- and y-intercepts. Plot these two points. Then draw a line through them.To find the x-intercept, let y = 0. �

� To find the y-intercept, let x = 0. �

So, the x-intercept is ±1 and the y-intercept is .

�

���±3x = 5 ± y

62/87,21���To graph the equation, find the x- and y-intercepts. Plot these two points. Then draw a line through them.To find the x-intercept, let y = 0. �

� To find the y-intercept, let x = 0. �

�

So, the x-intercept is �DQG�WKH�y-intercept is 5.

�

�

Solve each equation by graphing.���4x + 2 = 0

62/87,21���The related function is y = 4x + 2. 7R�JUDSK�WKH�IXQFWLRQ��PDNH�D�WDEOH���

The graph intersects the x-axis at . So, the solution is .

x f (x ) = 4x + 2 f (x �� (x, f(x ) ) ±4 f (±4) = 4(±4) + 2 ±14 (±4, ±14) ±2 f (±2) = 4(±2) + 2 ±6 (±2, ±6) ±0.5 f(±0.5) = 4(±0.5) + 2 0 (±0.5, 0)

0 f (0) = 4(0) + 2 2 (0, 2) 2 f (2) = 4(2) + 2 10 (2, 10) 4 f (4) = 4(4) + 2 18 (4, 18)

���0 = 6 ± 3x

62/87,21���The related function is y = í3x + 6. 7R�JUDSK�WKH�IXQFWLRQ��PDNH�D�WDEOH���

�

The graph intersects the x-axis at 2. So, the solution is 2.

x f (x ) = ±3x + 6 f (x �� (x, f(x ) ) ±4 f (±4) = ±3(±4) + 6 18 (±4, 18) ±2 f (±2) = ±3(±2) + 6 12 (±2, 12) 0 f (0) = ±3(0) + 6 6 (0, 6) 2 f (2) = ±3(2) + 6 0 (2, 0) 3 f (3) = ±3(3) + 6 ±3 (3, ±3) 4 f (4) = ±3(4) + 6 ±6 (4, ±6)

���5x + 2 = ±3

62/87,21���The related function is y = 5x + 5. 7R�JUDSK�WKH�IXQFWLRQ��PDNH�D�WDEOH���

�

The graph intersects the x-axis at ±1. So, the solution is ±1.

x f (x ) = 5x + 5 f (x �� (x, f(x ) ) ±4 f (±4) = 5(±4) + 5 ±15 (±4, ±15) ±2 f (±2) = 5(±2) + 5 ±5 (±2, ±5) ±1 f (±1) = 5(±1) + 5 0 (±1, 0) 0 f (0) = 5(0) + 5 5 (0, 5) 2 f (2) = 5(2) + 5 15 (2, 15) 4 f (4) = 5(4) + 5 25 (4, 25)

���12x = 4x + 16

62/87,21���The related function is y = í8x��������7R�JUDSK�WKH�IXQFWLRQ��PDNH�D�WDEOH���

�

The graph intersects the x-axis at 2. So, the solution is 2.

x f (x ) = ±8x + 16 f (x �� (x, f(x ) ) ±4 f (±4) = ±8(±4) + 16 48 (±4, 48) ±2 f (±2) = ±8(±2) + 16 32 (±2, 32) 0 f (0) = ±8(0) + 16 16 (0, 16) 2 f (2) = ±8(2) + 16 0 (2, 0) 3 f (3) = ±8(3) + 16 ±8 (3, ±8) 4 f (4) = ±8(4) + 16 ±16 (4, ±16)

Find the slope of the line that passes through each pair of points.����(5, 8), (±3, 7)

62/87,21���

So, the slope is .

����(5, ±2), (3, ±2)

62/87,21���

� So, the slope is 0.

����(±4, 7), (8, ±1)

62/87,21���

So, the slope is .

����(6, ±3), (6, 4)

62/87,21���

� So, the slope is undefined.

����MULTIPLE CHOICE Which is the slope of the linear function shown in the graph?�

A

B �

C

D

62/87,21���The line passes through the points (±2, 3) and (3, 1).�

The slope is , so the correct choice is B.

Suppose y varies directly as x. Write a direct variation equation that relates x and y . Then solve.����If y = 6 when x = 9, find x when y = 12.

62/87,21���

� So, the direct variation equation is ����Substitute 12 for y and find x.´��

� So, x = 18 when y = 12.

����When y = ±8, x = 8. What is x when y = ±6?

62/87,21���

� So, the direct variation equation is y = ±x���Substitute ±6 for y and find x.´��

� So, x = 6 when y = ±6.

����If y = ±5 when x = ±2, what is y when x = 14?

62/87,21���

� So, the direct variation equation is ���Substitute 14 for x and find y .�

� So, y = 35 when x = 14.

����If y = 2 when x = ±12, find y when x = ±4.

62/87,21���

� So, the direct variation equation is ����Substitute ±4 for x and find y .´��

�

So, y = �ZKHQ�x = ±4.

����BIOLOGY The number of pints of blood in a human body varies directly with the person¶s weight. A person who weighs 120 pounds has about 8.4 pints of blood in his or her body. a. Write and graph an equation relating weight and amount of blood in a person¶s body. b. Predict the weight of a person whose body holds 12 pints of blood.

62/87,21���a. To write a direct variation equation, find the constant of variation k . Let x = 120 and y = 8.4.�

� So, the direct variation equation is y = 0.07x.

b. Using the direct variation equation from part a, let y = 12. �

� So, a person who holds 12 pints of blood would weigh about 171 pounds.

Find the next three terms of each arithmetic sequence.����0, ±15, ±30, ±45, ±60, «

62/87,21���Find the common difference by subtracting two consecutive terms. � ±15 ± 0 = ±���� � The common difference between terms is ±�����6R��WR�ILQG�WKH�QH[W�WHUP��VXEWUDFW����IURP�WKH�ODVW�WHUP��7R�ILQG�WKH�next term, subtract 15 from the resulting number, and so on. � ±60 ± 15 = ±75 ±75 ± 15 = ±90 ±90 ± 15 = ±105 � So, the next three terms of this arithmetic sequence are ±75, ±90, ±105.

����5, 8, 11, 14, «

62/87,21���Find the common difference by subtracting two consecutive terms. � 8 ± �� ���� � The common difference between terms is 3. So, to find the next term, add 3 from the last term. To find the next term, add 3 from the resulting number, and so on. � 14 + 3 = 17 17 + 3 = 20 20 + 3 = 23 � So, the next three terms of this arithmetic sequence are 17, 20, 23.

Determine whether each sequence is an arithmetic sequence. If it is, state the common difference.����±40, ±32, ±24, ±16, «

62/87,21���±40, ±32, ±24, ±16, « � An arithmetic sequence is a numerical pattern that increases or decreases at a constant rate called the common difference. To find the common difference, subtract two consecutive numbers in the sequence. � ±32 ± (±40) = 8 ±24 ± (±32) = 8 ±16 ± (±24) = 7 � 7KH�GLIIHUHQFH�EHWZHHQ�WHUPV�LV�FRQVWDQW��VR�WKH�VHTXHQFH�LV�DQ�DULWKPHWLF�VHTXHQFH�� The common difference is 4. �

����0.75, 1.5, 3, 6, 12, «

62/87,21���An arithmetic sequence is a numerical pattern that increases or decreases at a constant rate called the common difference. To find the common difference, subtract two consecutive numbers in the sequence. � 1.5 ± 0.75 = 0.75 3 ± 1.5 = 1.5 6 ± 3 = 3 12 ± 6 = 6 � The difference between terms is not constant, so the sequence is not an an arithmetic sequence and does not have a common difference�� �

����5, 17, 29, 41, «

62/87,21���An arithmetic sequence is a numerical pattern that increases or decreases at a constant rate called the common difference. To find the common difference, subtract two consecutive numbers in the sequence. � 17 ± 5 = 12 29 ± 17 = 12 41 ± 29 = 12 � 7KH�GLIIHUHQFH�EHWZHHQ�WHUPV�LV�FRQVWDQW��VR�WKH�VHTXHQFH�LV�DQ�DULWKPHWLF�VHTXHQFH�� The common difference is 4. �

����MULTIPLE CHOICE In each figure, only one side of each regular pentagon is shared with another pentagon. Each side of each pentagon is 1 centimeter. If the pattern continues, what is the perimeter of a figure that has 6 pentagons?

F 15 cm H 20 cm G 25 cm J 30 cm

62/87,21���

� The perimeter increases by 3 as the number of polygons increases by 1. Continue the sequence to find the perimeter of a figure that has 6 pentagons. �

� So, a figure that has 6 pentagons has a perimeter of 20 centimeters. The correct choice is H.

Number of Pentagons

Perimeter

1 5 2 8 3 11

Number of Pentagons

Perimeter

4 14 5 17 6 20

eSolutions Manual - Powered by Cognero Page 4

Practice Test - Chapter 3

���TEMPERATURE The equation to convert Celsius temperature C to Kelvin temperature K is shown. �

� a. State the independent and dependent variables.Explain. b. Determine the C- and K-intercepts and describe what the intercepts mean in this situation.

62/87,21���a. The Kelvin temperature is dependent on the Celsius temperature. So, the Celsius temperature is the independent variable and the Kelvin temperature is the dependent variable. � b. The C-intercept is (±273, 0) and it means that a Celsius temperature of ±273 degrees is equal to a Kelvin temperature of 0 degrees. The K-intercept is (0, 273) and it means that a Celsius temperature of 0 degrees is equal to a Kelvin temperature of 273 degrees.

Graph each equation.���y = x + 2

62/87,21���To graph the equation, find the x- and y-intercepts. Plot these two points. Then draw a line through them.To find the x-intercept, let y = 0. �

� To find the y-intercept, let x = 0. �

� So, the x-intercept is ±2 and the y-intercept is 2. �

���y = 4x

62/87,21���The slope of y = 4x is 4. Graph (0, 0). From there, move up 4 units and right 1 unit to find another point. Then draw aline containing the points. �

���x + 2y = ±1

62/87,21���To graph the equation, find the x- and y-intercepts. Plot these two points. Then draw a line through them.To find the x-intercept, let y = 0. �

� To find the y-intercept, let x = 0. �

So, the x-intercept is ±1 and the y-intercept is .

�

���±3x = 5 ± y

62/87,21���To graph the equation, find the x- and y-intercepts. Plot these two points. Then draw a line through them.To find the x-intercept, let y = 0. �

� To find the y-intercept, let x = 0. �

�

So, the x-intercept is �DQG�WKH�y-intercept is 5.

�

�

Solve each equation by graphing.���4x + 2 = 0

62/87,21���The related function is y = 4x + 2. 7R�JUDSK�WKH�IXQFWLRQ��PDNH�D�WDEOH���

The graph intersects the x-axis at . So, the solution is .

x f (x ) = 4x + 2 f (x �� (x, f(x ) ) ±4 f (±4) = 4(±4) + 2 ±14 (±4, ±14) ±2 f (±2) = 4(±2) + 2 ±6 (±2, ±6) ±0.5 f(±0.5) = 4(±0.5) + 2 0 (±0.5, 0)

0 f (0) = 4(0) + 2 2 (0, 2) 2 f (2) = 4(2) + 2 10 (2, 10) 4 f (4) = 4(4) + 2 18 (4, 18)

���0 = 6 ± 3x

62/87,21���The related function is y = í3x + 6. 7R�JUDSK�WKH�IXQFWLRQ��PDNH�D�WDEOH���

�

The graph intersects the x-axis at 2. So, the solution is 2.

x f (x ) = ±3x + 6 f (x �� (x, f(x ) ) ±4 f (±4) = ±3(±4) + 6 18 (±4, 18) ±2 f (±2) = ±3(±2) + 6 12 (±2, 12) 0 f (0) = ±3(0) + 6 6 (0, 6) 2 f (2) = ±3(2) + 6 0 (2, 0) 3 f (3) = ±3(3) + 6 ±3 (3, ±3) 4 f (4) = ±3(4) + 6 ±6 (4, ±6)

���5x + 2 = ±3

62/87,21���The related function is y = 5x + 5. 7R�JUDSK�WKH�IXQFWLRQ��PDNH�D�WDEOH���

�

The graph intersects the x-axis at ±1. So, the solution is ±1.

x f (x ) = 5x + 5 f (x �� (x, f(x ) ) ±4 f (±4) = 5(±4) + 5 ±15 (±4, ±15) ±2 f (±2) = 5(±2) + 5 ±5 (±2, ±5) ±1 f (±1) = 5(±1) + 5 0 (±1, 0) 0 f (0) = 5(0) + 5 5 (0, 5) 2 f (2) = 5(2) + 5 15 (2, 15) 4 f (4) = 5(4) + 5 25 (4, 25)

���12x = 4x + 16

62/87,21���The related function is y = í8x��������7R�JUDSK�WKH�IXQFWLRQ��PDNH�D�WDEOH���

�

The graph intersects the x-axis at 2. So, the solution is 2.

x f (x ) = ±8x + 16 f (x �� (x, f(x ) ) ±4 f (±4) = ±8(±4) + 16 48 (±4, 48) ±2 f (±2) = ±8(±2) + 16 32 (±2, 32) 0 f (0) = ±8(0) + 16 16 (0, 16) 2 f (2) = ±8(2) + 16 0 (2, 0) 3 f (3) = ±8(3) + 16 ±8 (3, ±8) 4 f (4) = ±8(4) + 16 ±16 (4, ±16)

Find the slope of the line that passes through each pair of points.����(5, 8), (±3, 7)

62/87,21���

So, the slope is .

����(5, ±2), (3, ±2)

62/87,21���

� So, the slope is 0.

����(±4, 7), (8, ±1)

62/87,21���

So, the slope is .

����(6, ±3), (6, 4)

62/87,21���

� So, the slope is undefined.

����MULTIPLE CHOICE Which is the slope of the linear function shown in the graph?�

A

B �

C

D

62/87,21���The line passes through the points (±2, 3) and (3, 1).�

The slope is , so the correct choice is B.

Suppose y varies directly as x. Write a direct variation equation that relates x and y . Then solve.����If y = 6 when x = 9, find x when y = 12.

62/87,21���

� So, the direct variation equation is ����Substitute 12 for y and find x.´��

� So, x = 18 when y = 12.

����When y = ±8, x = 8. What is x when y = ±6?

62/87,21���

� So, the direct variation equation is y = ±x���Substitute ±6 for y and find x.´��

� So, x = 6 when y = ±6.

����If y = ±5 when x = ±2, what is y when x = 14?

62/87,21���

� So, the direct variation equation is ���Substitute 14 for x and find y .�

� So, y = 35 when x = 14.

����If y = 2 when x = ±12, find y when x = ±4.

62/87,21���

� So, the direct variation equation is ����Substitute ±4 for x and find y .´��

�

So, y = �ZKHQ�x = ±4.

����BIOLOGY The number of pints of blood in a human body varies directly with the person¶s weight. A person who weighs 120 pounds has about 8.4 pints of blood in his or her body. a. Write and graph an equation relating weight and amount of blood in a person¶s body. b. Predict the weight of a person whose body holds 12 pints of blood.

62/87,21���a. To write a direct variation equation, find the constant of variation k . Let x = 120 and y = 8.4.�

� So, the direct variation equation is y = 0.07x.

b. Using the direct variation equation from part a, let y = 12. �

� So, a person who holds 12 pints of blood would weigh about 171 pounds.

Find the next three terms of each arithmetic sequence.����0, ±15, ±30, ±45, ±60, «

62/87,21���Find the common difference by subtracting two consecutive terms. � ±15 ± 0 = ±���� � The common difference between terms is ±�����6R��WR�ILQG�WKH�QH[W�WHUP��VXEWUDFW����IURP�WKH�ODVW�WHUP��7R�ILQG�WKH�next term, subtract 15 from the resulting number, and so on. � ±60 ± 15 = ±75 ±75 ± 15 = ±90 ±90 ± 15 = ±105 � So, the next three terms of this arithmetic sequence are ±75, ±90, ±105.

����5, 8, 11, 14, «

62/87,21���Find the common difference by subtracting two consecutive terms. � 8 ± �� ���� � The common difference between terms is 3. So, to find the next term, add 3 from the last term. To find the next term, add 3 from the resulting number, and so on. � 14 + 3 = 17 17 + 3 = 20 20 + 3 = 23 � So, the next three terms of this arithmetic sequence are 17, 20, 23.

Determine whether each sequence is an arithmetic sequence. If it is, state the common difference.����±40, ±32, ±24, ±16, «

62/87,21���±40, ±32, ±24, ±16, « � An arithmetic sequence is a numerical pattern that increases or decreases at a constant rate called the common difference. To find the common difference, subtract two consecutive numbers in the sequence. � ±32 ± (±40) = 8 ±24 ± (±32) = 8 ±16 ± (±24) = 7 � 7KH�GLIIHUHQFH�EHWZHHQ�WHUPV�LV�FRQVWDQW��VR�WKH�VHTXHQFH�LV�DQ�DULWKPHWLF�VHTXHQFH�� The common difference is 4. �

����0.75, 1.5, 3, 6, 12, «

62/87,21���An arithmetic sequence is a numerical pattern that increases or decreases at a constant rate called the common difference. To find the common difference, subtract two consecutive numbers in the sequence. � 1.5 ± 0.75 = 0.75 3 ± 1.5 = 1.5 6 ± 3 = 3 12 ± 6 = 6 � The difference between terms is not constant, so the sequence is not an an arithmetic sequence and does not have a common difference�� �

����5, 17, 29, 41, «

62/87,21���An arithmetic sequence is a numerical pattern that increases or decreases at a constant rate called the common difference. To find the common difference, subtract two consecutive numbers in the sequence. � 17 ± 5 = 12 29 ± 17 = 12 41 ± 29 = 12 � 7KH�GLIIHUHQFH�EHWZHHQ�WHUPV�LV�FRQVWDQW��VR�WKH�VHTXHQFH�LV�DQ�DULWKPHWLF�VHTXHQFH�� The common difference is 4. �

����MULTIPLE CHOICE In each figure, only one side of each regular pentagon is shared with another pentagon. Each side of each pentagon is 1 centimeter. If the pattern continues, what is the perimeter of a figure that has 6 pentagons?

F 15 cm H 20 cm G 25 cm J 30 cm

62/87,21���

� The perimeter increases by 3 as the number of polygons increases by 1. Continue the sequence to find the perimeter of a figure that has 6 pentagons. �

� So, a figure that has 6 pentagons has a perimeter of 20 centimeters. The correct choice is H.

Number of Pentagons

Perimeter

1 5 2 8 3 11

Number of Pentagons

Perimeter

4 14 5 17 6 20

eSolutions Manual - Powered by Cognero Page 5

Practice Test - Chapter 3

���TEMPERATURE The equation to convert Celsius temperature C to Kelvin temperature K is shown. �

� a. State the independent and dependent variables.Explain. b. Determine the C- and K-intercepts and describe what the intercepts mean in this situation.

62/87,21���a. The Kelvin temperature is dependent on the Celsius temperature. So, the Celsius temperature is the independent variable and the Kelvin temperature is the dependent variable. � b. The C-intercept is (±273, 0) and it means that a Celsius temperature of ±273 degrees is equal to a Kelvin temperature of 0 degrees. The K-intercept is (0, 273) and it means that a Celsius temperature of 0 degrees is equal to a Kelvin temperature of 273 degrees.

Graph each equation.���y = x + 2

62/87,21���To graph the equation, find the x- and y-intercepts. Plot these two points. Then draw a line through them.To find the x-intercept, let y = 0. �

� To find the y-intercept, let x = 0. �

� So, the x-intercept is ±2 and the y-intercept is 2. �

���y = 4x

62/87,21���The slope of y = 4x is 4. Graph (0, 0). From there, move up 4 units and right 1 unit to find another point. Then draw aline containing the points. �

���x + 2y = ±1

62/87,21���To graph the equation, find the x- and y-intercepts. Plot these two points. Then draw a line through them.To find the x-intercept, let y = 0. �

� To find the y-intercept, let x = 0. �

So, the x-intercept is ±1 and the y-intercept is .

�

���±3x = 5 ± y

62/87,21���To graph the equation, find the x- and y-intercepts. Plot these two points. Then draw a line through them.To find the x-intercept, let y = 0. �

� To find the y-intercept, let x = 0. �

�

So, the x-intercept is �DQG�WKH�y-intercept is 5.

�

�

Solve each equation by graphing.���4x + 2 = 0

62/87,21���The related function is y = 4x + 2. 7R�JUDSK�WKH�IXQFWLRQ��PDNH�D�WDEOH���

The graph intersects the x-axis at . So, the solution is .

x f (x ) = 4x + 2 f (x �� (x, f(x ) ) ±4 f (±4) = 4(±4) + 2 ±14 (±4, ±14) ±2 f (±2) = 4(±2) + 2 ±6 (±2, ±6) ±0.5 f(±0.5) = 4(±0.5) + 2 0 (±0.5, 0)

0 f (0) = 4(0) + 2 2 (0, 2) 2 f (2) = 4(2) + 2 10 (2, 10) 4 f (4) = 4(4) + 2 18 (4, 18)

���0 = 6 ± 3x

62/87,21���The related function is y = í3x + 6. 7R�JUDSK�WKH�IXQFWLRQ��PDNH�D�WDEOH���

�

The graph intersects the x-axis at 2. So, the solution is 2.

x f (x ) = ±3x + 6 f (x �� (x, f(x ) ) ±4 f (±4) = ±3(±4) + 6 18 (±4, 18) ±2 f (±2) = ±3(±2) + 6 12 (±2, 12) 0 f (0) = ±3(0) + 6 6 (0, 6) 2 f (2) = ±3(2) + 6 0 (2, 0) 3 f (3) = ±3(3) + 6 ±3 (3, ±3) 4 f (4) = ±3(4) + 6 ±6 (4, ±6)

���5x + 2 = ±3

62/87,21���The related function is y = 5x + 5. 7R�JUDSK�WKH�IXQFWLRQ��PDNH�D�WDEOH���

�

The graph intersects the x-axis at ±1. So, the solution is ±1.

x f (x ) = 5x + 5 f (x �� (x, f(x ) ) ±4 f (±4) = 5(±4) + 5 ±15 (±4, ±15) ±2 f (±2) = 5(±2) + 5 ±5 (±2, ±5) ±1 f (±1) = 5(±1) + 5 0 (±1, 0) 0 f (0) = 5(0) + 5 5 (0, 5) 2 f (2) = 5(2) + 5 15 (2, 15) 4 f (4) = 5(4) + 5 25 (4, 25)

���12x = 4x + 16

62/87,21���The related function is y = í8x��������7R�JUDSK�WKH�IXQFWLRQ��PDNH�D�WDEOH���

�

The graph intersects the x-axis at 2. So, the solution is 2.

x f (x ) = ±8x + 16 f (x �� (x, f(x ) ) ±4 f (±4) = ±8(±4) + 16 48 (±4, 48) ±2 f (±2) = ±8(±2) + 16 32 (±2, 32) 0 f (0) = ±8(0) + 16 16 (0, 16) 2 f (2) = ±8(2) + 16 0 (2, 0) 3 f (3) = ±8(3) + 16 ±8 (3, ±8) 4 f (4) = ±8(4) + 16 ±16 (4, ±16)

Find the slope of the line that passes through each pair of points.����(5, 8), (±3, 7)

62/87,21���

So, the slope is .

����(5, ±2), (3, ±2)

62/87,21���

� So, the slope is 0.

����(±4, 7), (8, ±1)

62/87,21���

So, the slope is .

����(6, ±3), (6, 4)

62/87,21���

� So, the slope is undefined.

����MULTIPLE CHOICE Which is the slope of the linear function shown in the graph?�

A

B �

C

D

62/87,21���The line passes through the points (±2, 3) and (3, 1).�

The slope is , so the correct choice is B.

Suppose y varies directly as x. Write a direct variation equation that relates x and y . Then solve.����If y = 6 when x = 9, find x when y = 12.

62/87,21���

� So, the direct variation equation is ����Substitute 12 for y and find x.´��

� So, x = 18 when y = 12.

����When y = ±8, x = 8. What is x when y = ±6?

62/87,21���

� So, the direct variation equation is y = ±x���Substitute ±6 for y and find x.´��

� So, x = 6 when y = ±6.

����If y = ±5 when x = ±2, what is y when x = 14?

62/87,21���

� So, the direct variation equation is ���Substitute 14 for x and find y .�

� So, y = 35 when x = 14.

����If y = 2 when x = ±12, find y when x = ±4.

62/87,21���

� So, the direct variation equation is ����Substitute ±4 for x and find y .´��

�

So, y = �ZKHQ�x = ±4.

����BIOLOGY The number of pints of blood in a human body varies directly with the person¶s weight. A person who weighs 120 pounds has about 8.4 pints of blood in his or her body. a. Write and graph an equation relating weight and amount of blood in a person¶s body. b. Predict the weight of a person whose body holds 12 pints of blood.

62/87,21���a. To write a direct variation equation, find the constant of variation k . Let x = 120 and y = 8.4.�

� So, the direct variation equation is y = 0.07x.

b. Using the direct variation equation from part a, let y = 12. �

� So, a person who holds 12 pints of blood would weigh about 171 pounds.

Find the next three terms of each arithmetic sequence.����0, ±15, ±30, ±45, ±60, «

62/87,21���Find the common difference by subtracting two consecutive terms. � ±15 ± 0 = ±���� � The common difference between terms is ±�����6R��WR�ILQG�WKH�QH[W�WHUP��VXEWUDFW����IURP�WKH�ODVW�WHUP��7R�ILQG�WKH�next term, subtract 15 from the resulting number, and so on. � ±60 ± 15 = ±75 ±75 ± 15 = ±90 ±90 ± 15 = ±105 � So, the next three terms of this arithmetic sequence are ±75, ±90, ±105.

����5, 8, 11, 14, «

62/87,21���Find the common difference by subtracting two consecutive terms. � 8 ± �� ���� � The common difference between terms is 3. So, to find the next term, add 3 from the last term. To find the next term, add 3 from the resulting number, and so on. � 14 + 3 = 17 17 + 3 = 20 20 + 3 = 23 � So, the next three terms of this arithmetic sequence are 17, 20, 23.

Determine whether each sequence is an arithmetic sequence. If it is, state the common difference.����±40, ±32, ±24, ±16, «

62/87,21���±40, ±32, ±24, ±16, « � An arithmetic sequence is a numerical pattern that increases or decreases at a constant rate called the common difference. To find the common difference, subtract two consecutive numbers in the sequence. � ±32 ± (±40) = 8 ±24 ± (±32) = 8 ±16 ± (±24) = 7 � 7KH�GLIIHUHQFH�EHWZHHQ�WHUPV�LV�FRQVWDQW��VR�WKH�VHTXHQFH�LV�DQ�DULWKPHWLF�VHTXHQFH�� The common difference is 4. �

����0.75, 1.5, 3, 6, 12, «

62/87,21���An arithmetic sequence is a numerical pattern that increases or decreases at a constant rate called the common difference. To find the common difference, subtract two consecutive numbers in the sequence. � 1.5 ± 0.75 = 0.75 3 ± 1.5 = 1.5 6 ± 3 = 3 12 ± 6 = 6 � The difference between terms is not constant, so the sequence is not an an arithmetic sequence and does not have a common difference�� �

����5, 17, 29, 41, «

62/87,21���An arithmetic sequence is a numerical pattern that increases or decreases at a constant rate called the common difference. To find the common difference, subtract two consecutive numbers in the sequence. � 17 ± 5 = 12 29 ± 17 = 12 41 ± 29 = 12 � 7KH�GLIIHUHQFH�EHWZHHQ�WHUPV�LV�FRQVWDQW��VR�WKH�VHTXHQFH�LV�DQ�DULWKPHWLF�VHTXHQFH�� The common difference is 4. �

����MULTIPLE CHOICE In each figure, only one side of each regular pentagon is shared with another pentagon. Each side of each pentagon is 1 centimeter. If the pattern continues, what is the perimeter of a figure that has 6 pentagons?

F 15 cm H 20 cm G 25 cm J 30 cm

62/87,21���

� The perimeter increases by 3 as the number of polygons increases by 1. Continue the sequence to find the perimeter of a figure that has 6 pentagons. �

� So, a figure that has 6 pentagons has a perimeter of 20 centimeters. The correct choice is H.

Number of Pentagons

Perimeter

1 5 2 8 3 11

Number of Pentagons

Perimeter

4 14 5 17 6 20

eSolutions Manual - Powered by Cognero Page 6

Practice Test - Chapter 3

���TEMPERATURE The equation to convert Celsius temperature C to Kelvin temperature K is shown. �

� a. State the independent and dependent variables.Explain. b. Determine the C- and K-intercepts and describe what the intercepts mean in this situation.

62/87,21���a. The Kelvin temperature is dependent on the Celsius temperature. So, the Celsius temperature is the independent variable and the Kelvin temperature is the dependent variable. � b. The C-intercept is (±273, 0) and it means that a Celsius temperature of ±273 degrees is equal to a Kelvin temperature of 0 degrees. The K-intercept is (0, 273) and it means that a Celsius temperature of 0 degrees is equal to a Kelvin temperature of 273 degrees.

Graph each equation.���y = x + 2

62/87,21���To graph the equation, find the x- and y-intercepts. Plot these two points. Then draw a line through them.To find the x-intercept, let y = 0. �

� To find the y-intercept, let x = 0. �

� So, the x-intercept is ±2 and the y-intercept is 2. �

���y = 4x

62/87,21���The slope of y = 4x is 4. Graph (0, 0). From there, move up 4 units and right 1 unit to find another point. Then draw aline containing the points. �

���x + 2y = ±1

62/87,21���To graph the equation, find the x- and y-intercepts. Plot these two points. Then draw a line through them.To find the x-intercept, let y = 0. �

� To find the y-intercept, let x = 0. �

So, the x-intercept is ±1 and the y-intercept is .

�

���±3x = 5 ± y

62/87,21���To graph the equation, find the x- and y-intercepts. Plot these two points. Then draw a line through them.To find the x-intercept, let y = 0. �

� To find the y-intercept, let x = 0. �

�

So, the x-intercept is �DQG�WKH�y-intercept is 5.

�

�

Solve each equation by graphing.���4x + 2 = 0

62/87,21���The related function is y = 4x + 2. 7R�JUDSK�WKH�IXQFWLRQ��PDNH�D�WDEOH���

The graph intersects the x-axis at . So, the solution is .

x f (x ) = 4x + 2 f (x �� (x, f(x ) ) ±4 f (±4) = 4(±4) + 2 ±14 (±4, ±14) ±2 f (±2) = 4(±2) + 2 ±6 (±2, ±6) ±0.5 f(±0.5) = 4(±0.5) + 2 0 (±0.5, 0)

0 f (0) = 4(0) + 2 2 (0, 2) 2 f (2) = 4(2) + 2 10 (2, 10) 4 f (4) = 4(4) + 2 18 (4, 18)

���0 = 6 ± 3x

62/87,21���The related function is y = í3x + 6. 7R�JUDSK�WKH�IXQFWLRQ��PDNH�D�WDEOH���

�

The graph intersects the x-axis at 2. So, the solution is 2.

x f (x ) = ±3x + 6 f (x �� (x, f(x ) ) ±4 f (±4) = ±3(±4) + 6 18 (±4, 18) ±2 f (±2) = ±3(±2) + 6 12 (±2, 12) 0 f (0) = ±3(0) + 6 6 (0, 6) 2 f (2) = ±3(2) + 6 0 (2, 0) 3 f (3) = ±3(3) + 6 ±3 (3, ±3) 4 f (4) = ±3(4) + 6 ±6 (4, ±6)

���5x + 2 = ±3

62/87,21���The related function is y = 5x + 5. 7R�JUDSK�WKH�IXQFWLRQ��PDNH�D�WDEOH���

�

The graph intersects the x-axis at ±1. So, the solution is ±1.

x f (x ) = 5x + 5 f (x �� (x, f(x ) ) ±4 f (±4) = 5(±4) + 5 ±15 (±4, ±15) ±2 f (±2) = 5(±2) + 5 ±5 (±2, ±5) ±1 f (±1) = 5(±1) + 5 0 (±1, 0) 0 f (0) = 5(0) + 5 5 (0, 5) 2 f (2) = 5(2) + 5 15 (2, 15) 4 f (4) = 5(4) + 5 25 (4, 25)

���12x = 4x + 16

62/87,21���The related function is y = í8x��������7R�JUDSK�WKH�IXQFWLRQ��PDNH�D�WDEOH���

�

The graph intersects the x-axis at 2. So, the solution is 2.

x f (x ) = ±8x + 16 f (x �� (x, f(x ) ) ±4 f (±4) = ±8(±4) + 16 48 (±4, 48) ±2 f (±2) = ±8(±2) + 16 32 (±2, 32) 0 f (0) = ±8(0) + 16 16 (0, 16) 2 f (2) = ±8(2) + 16 0 (2, 0) 3 f (3) = ±8(3) + 16 ±8 (3, ±8) 4 f (4) = ±8(4) + 16 ±16 (4, ±16)

Find the slope of the line that passes through each pair of points.����(5, 8), (±3, 7)

62/87,21���

So, the slope is .

����(5, ±2), (3, ±2)

62/87,21���

� So, the slope is 0.

����(±4, 7), (8, ±1)

62/87,21���

So, the slope is .

����(6, ±3), (6, 4)

62/87,21���

� So, the slope is undefined.

����MULTIPLE CHOICE Which is the slope of the linear function shown in the graph?�

A

B �

C

D

62/87,21���The line passes through the points (±2, 3) and (3, 1).�

The slope is , so the correct choice is B.

Suppose y varies directly as x. Write a direct variation equation that relates x and y . Then solve.����If y = 6 when x = 9, find x when y = 12.

62/87,21���

� So, the direct variation equation is ����Substitute 12 for y and find x.´��

� So, x = 18 when y = 12.

����When y = ±8, x = 8. What is x when y = ±6?

62/87,21���

� So, the direct variation equation is y = ±x���Substitute ±6 for y and find x.´��

� So, x = 6 when y = ±6.

����If y = ±5 when x = ±2, what is y when x = 14?

62/87,21���

� So, the direct variation equation is ���Substitute 14 for x and find y .�

� So, y = 35 when x = 14.

����If y = 2 when x = ±12, find y when x = ±4.

62/87,21���

� So, the direct variation equation is ����Substitute ±4 for x and find y .´��

�

So, y = �ZKHQ�x = ±4.

����BIOLOGY The number of pints of blood in a human body varies directly with the person¶s weight. A person who weighs 120 pounds has about 8.4 pints of blood in his or her body. a. Write and graph an equation relating weight and amount of blood in a person¶s body. b. Predict the weight of a person whose body holds 12 pints of blood.

62/87,21���a. To write a direct variation equation, find the constant of variation k . Let x = 120 and y = 8.4.�

� So, the direct variation equation is y = 0.07x.

b. Using the direct variation equation from part a, let y = 12. �

� So, a person who holds 12 pints of blood would weigh about 171 pounds.

Find the next three terms of each arithmetic sequence.����0, ±15, ±30, ±45, ±60, «

62/87,21���Find the common difference by subtracting two consecutive terms. � ±15 ± 0 = ±���� � The common difference between terms is ±�����6R��WR�ILQG�WKH�QH[W�WHUP��VXEWUDFW����IURP�WKH�ODVW�WHUP��7R�ILQG�WKH�next term, subtract 15 from the resulting number, and so on. � ±60 ± 15 = ±75 ±75 ± 15 = ±90 ±90 ± 15 = ±105 � So, the next three terms of this arithmetic sequence are ±75, ±90, ±105.

����5, 8, 11, 14, «

62/87,21���Find the common difference by subtracting two consecutive terms. � 8 ± �� ���� � The common difference between terms is 3. So, to find the next term, add 3 from the last term. To find the next term, add 3 from the resulting number, and so on. � 14 + 3 = 17 17 + 3 = 20 20 + 3 = 23 � So, the next three terms of this arithmetic sequence are 17, 20, 23.

Determine whether each sequence is an arithmetic sequence. If it is, state the common difference.����±40, ±32, ±24, ±16, «

62/87,21���±40, ±32, ±24, ±16, « � An arithmetic sequence is a numerical pattern that increases or decreases at a constant rate called the common difference. To find the common difference, subtract two consecutive numbers in the sequence. � ±32 ± (±40) = 8 ±24 ± (±32) = 8 ±16 ± (±24) = 7 � 7KH�GLIIHUHQFH�EHWZHHQ�WHUPV�LV�FRQVWDQW��VR�WKH�VHTXHQFH�LV�DQ�DULWKPHWLF�VHTXHQFH�� The common difference is 4. �

����0.75, 1.5, 3, 6, 12, «

62/87,21���An arithmetic sequence is a numerical pattern that increases or decreases at a constant rate called the common difference. To find the common difference, subtract two consecutive numbers in the sequence. � 1.5 ± 0.75 = 0.75 3 ± 1.5 = 1.5 6 ± 3 = 3 12 ± 6 = 6 � The difference between terms is not constant, so the sequence is not an an arithmetic sequence and does not have a common difference�� �

����5, 17, 29, 41, «

62/87,21���An arithmetic sequence is a numerical pattern that increases or decreases at a constant rate called the common difference. To find the common difference, subtract two consecutive numbers in the sequence. � 17 ± 5 = 12 29 ± 17 = 12 41 ± 29 = 12 � 7KH�GLIIHUHQFH�EHWZHHQ�WHUPV�LV�FRQVWDQW��VR�WKH�VHTXHQFH�LV�DQ�DULWKPHWLF�VHTXHQFH�� The common difference is 4. �

����MULTIPLE CHOICE In each figure, only one side of each regular pentagon is shared with another pentagon. Each side of each pentagon is 1 centimeter. If the pattern continues, what is the perimeter of a figure that has 6 pentagons?

F 15 cm H 20 cm G 25 cm J 30 cm

62/87,21���

� The perimeter increases by 3 as the number of polygons increases by 1. Continue the sequence to find the perimeter of a figure that has 6 pentagons. �

� So, a figure that has 6 pentagons has a perimeter of 20 centimeters. The correct choice is H.

Number of Pentagons

Perimeter

1 5 2 8 3 11

Number of Pentagons

Perimeter

4 14 5 17 6 20

eSolutions Manual - Powered by Cognero Page 7

Practice Test - Chapter 3

���TEMPERATURE The equation to convert Celsius temperature C to Kelvin temperature K is shown. �

� a. State the independent and dependent variables.Explain. b. Determine the C- and K-intercepts and describe what the intercepts mean in this situation.

62/87,21���a. The Kelvin temperature is dependent on the Celsius temperature. So, the Celsius temperature is the independent variable and the Kelvin temperature is the dependent variable. � b. The C-intercept is (±273, 0) and it means that a Celsius temperature of ±273 degrees is equal to a Kelvin temperature of 0 degrees. The K-intercept is (0, 273) and it means that a Celsius temperature of 0 degrees is equal to a Kelvin temperature of 273 degrees.

Graph each equation.���y = x + 2

62/87,21���To graph the equation, find the x- and y-intercepts. Plot these two points. Then draw a line through them.To find the x-intercept, let y = 0. �

� To find the y-intercept, let x = 0. �

� So, the x-intercept is ±2 and the y-intercept is 2. �

���y = 4x

62/87,21���The slope of y = 4x is 4. Graph (0, 0). From there, move up 4 units and right 1 unit to find another point. Then draw aline containing the points. �

���x + 2y = ±1

62/87,21���To graph the equation, find the x- and y-intercepts. Plot these two points. Then draw a line through them.To find the x-intercept, let y = 0. �

� To find the y-intercept, let x = 0. �

So, the x-intercept is ±1 and the y-intercept is .

�

���±3x = 5 ± y

62/87,21���To graph the equation, find the x- and y-intercepts. Plot these two points. Then draw a line through them.To find the x-intercept, let y = 0. �

� To find the y-intercept, let x = 0. �

�

So, the x-intercept is �DQG�WKH�y-intercept is 5.

�

�

Solve each equation by graphing.���4x + 2 = 0

62/87,21���The related function is y = 4x + 2. 7R�JUDSK�WKH�IXQFWLRQ��PDNH�D�WDEOH���

The graph intersects the x-axis at . So, the solution is .

x f (x ) = 4x + 2 f (x �� (x, f(x ) ) ±4 f (±4) = 4(±4) + 2 ±14 (±4, ±14) ±2 f (±2) = 4(±2) + 2 ±6 (±2, ±6) ±0.5 f(±0.5) = 4(±0.5) + 2 0 (±0.5, 0)

0 f (0) = 4(0) + 2 2 (0, 2) 2 f (2) = 4(2) + 2 10 (2, 10) 4 f (4) = 4(4) + 2 18 (4, 18)

���0 = 6 ± 3x

62/87,21���The related function is y = í3x + 6. 7R�JUDSK�WKH�IXQFWLRQ��PDNH�D�WDEOH���

�

The graph intersects the x-axis at 2. So, the solution is 2.

x f (x ) = ±3x + 6 f (x �� (x, f(x ) ) ±4 f (±4) = ±3(±4) + 6 18 (±4, 18) ±2 f (±2) = ±3(±2) + 6 12 (±2, 12) 0 f (0) = ±3(0) + 6 6 (0, 6) 2 f (2) = ±3(2) + 6 0 (2, 0) 3 f (3) = ±3(3) + 6 ±3 (3, ±3) 4 f (4) = ±3(4) + 6 ±6 (4, ±6)

���5x + 2 = ±3

62/87,21���The related function is y = 5x + 5. 7R�JUDSK�WKH�IXQFWLRQ��PDNH�D�WDEOH���

�

The graph intersects the x-axis at ±1. So, the solution is ±1.

x f (x ) = 5x + 5 f (x �� (x, f(x ) ) ±4 f (±4) = 5(±4) + 5 ±15 (±4, ±15) ±2 f (±2) = 5(±2) + 5 ±5 (±2, ±5) ±1 f (±1) = 5(±1) + 5 0 (±1, 0) 0 f (0) = 5(0) + 5 5 (0, 5) 2 f (2) = 5(2) + 5 15 (2, 15) 4 f (4) = 5(4) + 5 25 (4, 25)

���12x = 4x + 16

62/87,21���The related function is y = í8x��������7R�JUDSK�WKH�IXQFWLRQ��PDNH�D�WDEOH���

�

The graph intersects the x-axis at 2. So, the solution is 2.

x f (x ) = ±8x + 16 f (x �� (x, f(x ) ) ±4 f (±4) = ±8(±4) + 16 48 (±4, 48) ±2 f (±2) = ±8(±2) + 16 32 (±2, 32) 0 f (0) = ±8(0) + 16 16 (0, 16) 2 f (2) = ±8(2) + 16 0 (2, 0) 3 f (3) = ±8(3) + 16 ±8 (3, ±8) 4 f (4) = ±8(4) + 16 ±16 (4, ±16)

Find the slope of the line that passes through each pair of points.����(5, 8), (±3, 7)

62/87,21���

So, the slope is .

����(5, ±2), (3, ±2)

62/87,21���

� So, the slope is 0.

����(±4, 7), (8, ±1)

62/87,21���

So, the slope is .

����(6, ±3), (6, 4)

62/87,21���

� So, the slope is undefined.

����MULTIPLE CHOICE Which is the slope of the linear function shown in the graph?�

A

B �

C

D

62/87,21���The line passes through the points (±2, 3) and (3, 1).�

The slope is , so the correct choice is B.

Suppose y varies directly as x. Write a direct variation equation that relates x and y . Then solve.����If y = 6 when x = 9, find x when y = 12.

62/87,21���

� So, the direct variation equation is ����Substitute 12 for y and find x.´��

� So, x = 18 when y = 12.

����When y = ±8, x = 8. What is x when y = ±6?

62/87,21���

� So, the direct variation equation is y = ±x���Substitute ±6 for y and find x.´��

� So, x = 6 when y = ±6.

����If y = ±5 when x = ±2, what is y when x = 14?

62/87,21���

� So, the direct variation equation is ���Substitute 14 for x and find y .�

� So, y = 35 when x = 14.

����If y = 2 when x = ±12, find y when x = ±4.

62/87,21���

� So, the direct variation equation is ����Substitute ±4 for x and find y .´��

�

So, y = �ZKHQ�x = ±4.

����BIOLOGY The number of pints of blood in a human body varies directly with the person¶s weight. A person who weighs 120 pounds has about 8.4 pints of blood in his or her body. a. Write and graph an equation relating weight and amount of blood in a person¶s body. b. Predict the weight of a person whose body holds 12 pints of blood.

62/87,21���a. To write a direct variation equation, find the constant of variation k . Let x = 120 and y = 8.4.�

� So, the direct variation equation is y = 0.07x.

b. Using the direct variation equation from part a, let y = 12. �

� So, a person who holds 12 pints of blood would weigh about 171 pounds.

Find the next three terms of each arithmetic sequence.����0, ±15, ±30, ±45, ±60, «

62/87,21���Find the common difference by subtracting two consecutive terms. � ±15 ± 0 = ±���� � The common difference between terms is ±�����6R��WR�ILQG�WKH�QH[W�WHUP��VXEWUDFW����IURP�WKH�ODVW�WHUP��7R�ILQG�WKH�next term, subtract 15 from the resulting number, and so on. � ±60 ± 15 = ±75 ±75 ± 15 = ±90 ±90 ± 15 = ±105 � So, the next three terms of this arithmetic sequence are ±75, ±90, ±105.

����5, 8, 11, 14, «

62/87,21���Find the common difference by subtracting two consecutive terms. � 8 ± �� ���� � The common difference between terms is 3. So, to find the next term, add 3 from the last term. To find the next term, add 3 from the resulting number, and so on. � 14 + 3 = 17 17 + 3 = 20 20 + 3 = 23 � So, the next three terms of this arithmetic sequence are 17, 20, 23.

Determine whether each sequence is an arithmetic sequence. If it is, state the common difference.����±40, ±32, ±24, ±16, «

62/87,21���±40, ±32, ±24, ±16, « � An arithmetic sequence is a numerical pattern that increases or decreases at a constant rate called the common difference. To find the common difference, subtract two consecutive numbers in the sequence. � ±32 ± (±40) = 8 ±24 ± (±32) = 8 ±16 ± (±24) = 7 � 7KH�GLIIHUHQFH�EHWZHHQ�WHUPV�LV�FRQVWDQW��VR�WKH�VHTXHQFH�LV�DQ�DULWKPHWLF�VHTXHQFH�� The common difference is 4. �

����0.75, 1.5, 3, 6, 12, «

62/87,21���An arithmetic sequence is a numerical pattern that increases or decreases at a constant rate called the common difference. To find the common difference, subtract two consecutive numbers in the sequence. � 1.5 ± 0.75 = 0.75 3 ± 1.5 = 1.5 6 ± 3 = 3 12 ± 6 = 6 � The difference between terms is not constant, so the sequence is not an an arithmetic sequence and does not have a common difference�� �

����5, 17, 29, 41, «

62/87,21���An arithmetic sequence is a numerical pattern that increases or decreases at a constant rate called the common difference. To find the common difference, subtract two consecutive numbers in the sequence. � 17 ± 5 = 12 29 ± 17 = 12 41 ± 29 = 12 � 7KH�GLIIHUHQFH�EHWZHHQ�WHUPV�LV�FRQVWDQW��VR�WKH�VHTXHQFH�LV�DQ�DULWKPHWLF�VHTXHQFH�� The common difference is 4. �

����MULTIPLE CHOICE In each figure, only one side of each regular pentagon is shared with another pentagon. Each side of each pentagon is 1 centimeter. If the pattern continues, what is the perimeter of a figure that has 6 pentagons?

F 15 cm H 20 cm G 25 cm J 30 cm

62/87,21���

� The perimeter increases by 3 as the number of polygons increases by 1. Continue the sequence to find the perimeter of a figure that has 6 pentagons. �

� So, a figure that has 6 pentagons has a perimeter of 20 centimeters. The correct choice is H.

Number of Pentagons

Perimeter

1 5 2 8 3 11

Number of Pentagons

Perimeter

4 14 5 17 6 20

eSolutions Manual - Powered by Cognero Page 8

Practice Test - Chapter 3

���TEMPERATURE The equation to convert Celsius temperature C to Kelvin temperature K is shown. �

� a. State the independent and dependent variables.Explain. b. Determine the C- and K-intercepts and describe what the intercepts mean in this situation.

62/87,21���a. The Kelvin temperature is dependent on the Celsius temperature. So, the Celsius temperature is the independent variable and the Kelvin temperature is the dependent variable. � b. The C-intercept is (±273, 0) and it means that a Celsius temperature of ±273 degrees is equal to a Kelvin temperature of 0 degrees. The K-intercept is (0, 273) and it means that a Celsius temperature of 0 degrees is equal to a Kelvin temperature of 273 degrees.

Graph each equation.���y = x + 2

62/87,21���To graph the equation, find the x- and y-intercepts. Plot these two points. Then draw a line through them.To find the x-intercept, let y = 0. �

� To find the y-intercept, let x = 0. �

� So, the x-intercept is ±2 and the y-intercept is 2. �

���y = 4x

62/87,21���The slope of y = 4x is 4. Graph (0, 0). From there, move up 4 units and right 1 unit to find another point. Then draw aline containing the points. �

���x + 2y = ±1

62/87,21���To graph the equation, find the x- and y-intercepts. Plot these two points. Then draw a line through them.To find the x-intercept, let y = 0. �

� To find the y-intercept, let x = 0. �

So, the x-intercept is ±1 and the y-intercept is .

�

���±3x = 5 ± y

62/87,21���To graph the equation, find the x- and y-intercepts. Plot these two points. Then draw a line through them.To find the x-intercept, let y = 0. �

� To find the y-intercept, let x = 0. �

�

So, the x-intercept is �DQG�WKH�y-intercept is 5.

�

�

Solve each equation by graphing.���4x + 2 = 0

62/87,21���The related function is y = 4x + 2. 7R�JUDSK�WKH�IXQFWLRQ��PDNH�D�WDEOH���

The graph intersects the x-axis at . So, the solution is .

x f (x ) = 4x + 2 f (x �� (x, f(x ) ) ±4 f (±4) = 4(±4) + 2 ±14 (±4, ±14) ±2 f (±2) = 4(±2) + 2 ±6 (±2, ±6) ±0.5 f(±0.5) = 4(±0.5) + 2 0 (±0.5, 0)

0 f (0) = 4(0) + 2 2 (0, 2) 2 f (2) = 4(2) + 2 10 (2, 10) 4 f (4) = 4(4) + 2 18 (4, 18)

���0 = 6 ± 3x

62/87,21���The related function is y = í3x + 6. 7R�JUDSK�WKH�IXQFWLRQ��PDNH�D�WDEOH���

�

The graph intersects the x-axis at 2. So, the solution is 2.

x f (x ) = ±3x + 6 f (x �� (x, f(x ) ) ±4 f (±4) = ±3(±4) + 6 18 (±4, 18) ±2 f (±2) = ±3(±2) + 6 12 (±2, 12) 0 f (0) = ±3(0) + 6 6 (0, 6) 2 f (2) = ±3(2) + 6 0 (2, 0) 3 f (3) = ±3(3) + 6 ±3 (3, ±3) 4 f (4) = ±3(4) + 6 ±6 (4, ±6)

���5x + 2 = ±3

62/87,21���The related function is y = 5x + 5. 7R�JUDSK�WKH�IXQFWLRQ��PDNH�D�WDEOH���

�

The graph intersects the x-axis at ±1. So, the solution is ±1.

x f (x ) = 5x + 5 f (x �� (x, f(x ) ) ±4 f (±4) = 5(±4) + 5 ±15 (±4, ±15) ±2 f (±2) = 5(±2) + 5 ±5 (±2, ±5) ±1 f (±1) = 5(±1) + 5 0 (±1, 0) 0 f (0) = 5(0) + 5 5 (0, 5) 2 f (2) = 5(2) + 5 15 (2, 15) 4 f (4) = 5(4) + 5 25 (4, 25)

���12x = 4x + 16

62/87,21���The related function is y = í8x��������7R�JUDSK�WKH�IXQFWLRQ��PDNH�D�WDEOH���

�

The graph intersects the x-axis at 2. So, the solution is 2.

x f (x ) = ±8x + 16 f (x �� (x, f(x ) ) ±4 f (±4) = ±8(±4) + 16 48 (±4, 48) ±2 f (±2) = ±8(±2) + 16 32 (±2, 32) 0 f (0) = ±8(0) + 16 16 (0, 16) 2 f (2) = ±8(2) + 16 0 (2, 0) 3 f (3) = ±8(3) + 16 ±8 (3, ±8) 4 f (4) = ±8(4) + 16 ±16 (4, ±16)

Find the slope of the line that passes through each pair of points.����(5, 8), (±3, 7)

62/87,21���

So, the slope is .

����(5, ±2), (3, ±2)

62/87,21���

� So, the slope is 0.

����(±4, 7), (8, ±1)

62/87,21���

So, the slope is .

����(6, ±3), (6, 4)

62/87,21���

� So, the slope is undefined.

����MULTIPLE CHOICE Which is the slope of the linear function shown in the graph?�

A

B �

C

D

62/87,21���The line passes through the points (±2, 3) and (3, 1).�

The slope is , so the correct choice is B.

Suppose y varies directly as x. Write a direct variation equation that relates x and y . Then solve.����If y = 6 when x = 9, find x when y = 12.

62/87,21���

� So, the direct variation equation is ����Substitute 12 for y and find x.´��

� So, x = 18 when y = 12.

����When y = ±8, x = 8. What is x when y = ±6?

62/87,21���

� So, the direct variation equation is y = ±x���Substitute ±6 for y and find x.´��

� So, x = 6 when y = ±6.

����If y = ±5 when x = ±2, what is y when x = 14?

62/87,21���

� So, the direct variation equation is ���Substitute 14 for x and find y .�

� So, y = 35 when x = 14.

����If y = 2 when x = ±12, find y when x = ±4.

62/87,21���

� So, the direct variation equation is ����Substitute ±4 for x and find y .´��

�

So, y = �ZKHQ�x = ±4.

����BIOLOGY The number of pints of blood in a human body varies directly with the person¶s weight. A person who weighs 120 pounds has about 8.4 pints of blood in his or her body. a. Write and graph an equation relating weight and amount of blood in a person¶s body. b. Predict the weight of a person whose body holds 12 pints of blood.

62/87,21���a. To write a direct variation equation, find the constant of variation k . Let x = 120 and y = 8.4.�

� So, the direct variation equation is y = 0.07x.

b. Using the direct variation equation from part a, let y = 12. �

� So, a person who holds 12 pints of blood would weigh about 171 pounds.

Find the next three terms of each arithmetic sequence.����0, ±15, ±30, ±45, ±60, «

62/87,21���Find the common difference by subtracting two consecutive terms. � ±15 ± 0 = ±���� � The common difference between terms is ±�����6R��WR�ILQG�WKH�QH[W�WHUP��VXEWUDFW����IURP�WKH�ODVW�WHUP��7R�ILQG�WKH�next term, subtract 15 from the resulting number, and so on. � ±60 ± 15 = ±75 ±75 ± 15 = ±90 ±90 ± 15 = ±105 � So, the next three terms of this arithmetic sequence are ±75, ±90, ±105.

����5, 8, 11, 14, «

62/87,21���Find the common difference by subtracting two consecutive terms. � 8 ± �� ���� � The common difference between terms is 3. So, to find the next term, add 3 from the last term. To find the next term, add 3 from the resulting number, and so on. � 14 + 3 = 17 17 + 3 = 20 20 + 3 = 23 � So, the next three terms of this arithmetic sequence are 17, 20, 23.

Determine whether each sequence is an arithmetic sequence. If it is, state the common difference.����±40, ±32, ±24, ±16, «

62/87,21���±40, ±32, ±24, ±16, « � An arithmetic sequence is a numerical pattern that increases or decreases at a constant rate called the common difference. To find the common difference, subtract two consecutive numbers in the sequence. � ±32 ± (±40) = 8 ±24 ± (±32) = 8 ±16 ± (±24) = 7 � 7KH�GLIIHUHQFH�EHWZHHQ�WHUPV�LV�FRQVWDQW��VR�WKH�VHTXHQFH�LV�DQ�DULWKPHWLF�VHTXHQFH�� The common difference is 4. �

����0.75, 1.5, 3, 6, 12, «

62/87,21���An arithmetic sequence is a numerical pattern that increases or decreases at a constant rate called the common difference. To find the common difference, subtract two consecutive numbers in the sequence. � 1.5 ± 0.75 = 0.75 3 ± 1.5 = 1.5 6 ± 3 = 3 12 ± 6 = 6 � The difference between terms is not constant, so the sequence is not an an arithmetic sequence and does not have a common difference�� �

����5, 17, 29, 41, «

62/87,21���An arithmetic sequence is a numerical pattern that increases or decreases at a constant rate called the common difference. To find the common difference, subtract two consecutive numbers in the sequence. � 17 ± 5 = 12 29 ± 17 = 12 41 ± 29 = 12 � 7KH�GLIIHUHQFH�EHWZHHQ�WHUPV�LV�FRQVWDQW��VR�WKH�VHTXHQFH�LV�DQ�DULWKPHWLF�VHTXHQFH�� The common difference is 4. �

����MULTIPLE CHOICE In each figure, only one side of each regular pentagon is shared with another pentagon. Each side of each pentagon is 1 centimeter. If the pattern continues, what is the perimeter of a figure that has 6 pentagons?

F 15 cm H 20 cm G 25 cm J 30 cm

62/87,21���

� The perimeter increases by 3 as the number of polygons increases by 1. Continue the sequence to find the perimeter of a figure that has 6 pentagons. �

� So, a figure that has 6 pentagons has a perimeter of 20 centimeters. The correct choice is H.

Number of Pentagons

Perimeter

1 5 2 8 3 11

Number of Pentagons

Perimeter

4 14 5 17 6 20

eSolutions Manual - Powered by Cognero Page 9

Practice Test - Chapter 3

���TEMPERATURE The equation to convert Celsius temperature C to Kelvin temperature K is shown. �

� a. State the independent and dependent variables.Explain. b. Determine the C- and K-intercepts and describe what the intercepts mean in this situation.

62/87,21���a. The Kelvin temperature is dependent on the Celsius temperature. So, the Celsius temperature is the independent variable and the Kelvin temperature is the dependent variable. � b. The C-intercept is (±273, 0) and it means that a Celsius temperature of ±273 degrees is equal to a Kelvin temperature of 0 degrees. The K-intercept is (0, 273) and it means that a Celsius temperature of 0 degrees is equal to a Kelvin temperature of 273 degrees.

Graph each equation.���y = x + 2

62/87,21���To graph the equation, find the x- and y-intercepts. Plot these two points. Then draw a line through them.To find the x-intercept, let y = 0. �

� To find the y-intercept, let x = 0. �

� So, the x-intercept is ±2 and the y-intercept is 2. �

���y = 4x

62/87,21���The slope of y = 4x is 4. Graph (0, 0). From there, move up 4 units and right 1 unit to find another point. Then draw aline containing the points. �

���x + 2y = ±1

62/87,21���To graph the equation, find the x- and y-intercepts. Plot these two points. Then draw a line through them.To find the x-intercept, let y = 0. �

� To find the y-intercept, let x = 0. �

So, the x-intercept is ±1 and the y-intercept is .

�

���±3x = 5 ± y

62/87,21���To graph the equation, find the x- and y-intercepts. Plot these two points. Then draw a line through them.To find the x-intercept, let y = 0. �

� To find the y-intercept, let x = 0. �

�

So, the x-intercept is �DQG�WKH�y-intercept is 5.

�

�

Solve each equation by graphing.���4x + 2 = 0

62/87,21���The related function is y = 4x + 2. 7R�JUDSK�WKH�IXQFWLRQ��PDNH�D�WDEOH���

The graph intersects the x-axis at . So, the solution is .

x f (x ) = 4x + 2 f (x �� (x, f(x ) ) ±4 f (±4) = 4(±4) + 2 ±14 (±4, ±14) ±2 f (±2) = 4(±2) + 2 ±6 (±2, ±6) ±0.5 f(±0.5) = 4(±0.5) + 2 0 (±0.5, 0)

0 f (0) = 4(0) + 2 2 (0, 2) 2 f (2) = 4(2) + 2 10 (2, 10) 4 f (4) = 4(4) + 2 18 (4, 18)

���0 = 6 ± 3x

62/87,21���The related function is y = í3x + 6. 7R�JUDSK�WKH�IXQFWLRQ��PDNH�D�WDEOH���

�

The graph intersects the x-axis at 2. So, the solution is 2.

x f (x ) = ±3x + 6 f (x �� (x, f(x ) ) ±4 f (±4) = ±3(±4) + 6 18 (±4, 18) ±2 f (±2) = ±3(±2) + 6 12 (±2, 12) 0 f (0) = ±3(0) + 6 6 (0, 6) 2 f (2) = ±3(2) + 6 0 (2, 0) 3 f (3) = ±3(3) + 6 ±3 (3, ±3) 4 f (4) = ±3(4) + 6 ±6 (4, ±6)

���5x + 2 = ±3

62/87,21���The related function is y = 5x + 5. 7R�JUDSK�WKH�IXQFWLRQ��PDNH�D�WDEOH���

�

The graph intersects the x-axis at ±1. So, the solution is ±1.

x f (x ) = 5x + 5 f (x �� (x, f(x ) ) ±4 f (±4) = 5(±4) + 5 ±15 (±4, ±15) ±2 f (±2) = 5(±2) + 5 ±5 (±2, ±5) ±1 f (±1) = 5(±1) + 5 0 (±1, 0) 0 f (0) = 5(0) + 5 5 (0, 5) 2 f (2) = 5(2) + 5 15 (2, 15) 4 f (4) = 5(4) + 5 25 (4, 25)

���12x = 4x + 16

62/87,21���The related function is y = í8x��������7R�JUDSK�WKH�IXQFWLRQ��PDNH�D�WDEOH���

�

The graph intersects the x-axis at 2. So, the solution is 2.

x f (x ) = ±8x + 16 f (x �� (x, f(x ) ) ±4 f (±4) = ±8(±4) + 16 48 (±4, 48) ±2 f (±2) = ±8(±2) + 16 32 (±2, 32) 0 f (0) = ±8(0) + 16 16 (0, 16) 2 f (2) = ±8(2) + 16 0 (2, 0) 3 f (3) = ±8(3) + 16 ±8 (3, ±8) 4 f (4) = ±8(4) + 16 ±16 (4, ±16)

Find the slope of the line that passes through each pair of points.����(5, 8), (±3, 7)

62/87,21���

So, the slope is .

����(5, ±2), (3, ±2)

62/87,21���

� So, the slope is 0.

����(±4, 7), (8, ±1)

62/87,21���

So, the slope is .

����(6, ±3), (6, 4)

62/87,21���

� So, the slope is undefined.

����MULTIPLE CHOICE Which is the slope of the linear function shown in the graph?�

A

B �

C

D

62/87,21���The line passes through the points (±2, 3) and (3, 1).�

The slope is , so the correct choice is B.

Suppose y varies directly as x. Write a direct variation equation that relates x and y . Then solve.����If y = 6 when x = 9, find x when y = 12.

62/87,21���

� So, the direct variation equation is ����Substitute 12 for y and find x.´��

� So, x = 18 when y = 12.

����When y = ±8, x = 8. What is x when y = ±6?

62/87,21���

� So, the direct variation equation is y = ±x���Substitute ±6 for y and find x.´��

� So, x = 6 when y = ±6.

����If y = ±5 when x = ±2, what is y when x = 14?

62/87,21���

� So, the direct variation equation is ���Substitute 14 for x and find y .�

� So, y = 35 when x = 14.

����If y = 2 when x = ±12, find y when x = ±4.

62/87,21���

� So, the direct variation equation is ����Substitute ±4 for x and find y .´��

�

So, y = �ZKHQ�x = ±4.

����BIOLOGY The number of pints of blood in a human body varies directly with the person¶s weight. A person who weighs 120 pounds has about 8.4 pints of blood in his or her body. a. Write and graph an equation relating weight and amount of blood in a person¶s body. b. Predict the weight of a person whose body holds 12 pints of blood.

62/87,21���a. To write a direct variation equation, find the constant of variation k . Let x = 120 and y = 8.4.�

� So, the direct variation equation is y = 0.07x.

b. Using the direct variation equation from part a, let y = 12. �

� So, a person who holds 12 pints of blood would weigh about 171 pounds.

Find the next three terms of each arithmetic sequence.����0, ±15, ±30, ±45, ±60, «

62/87,21���Find the common difference by subtracting two consecutive terms. � ±15 ± 0 = ±���� � The common difference between terms is ±�����6R��WR�ILQG�WKH�QH[W�WHUP��VXEWUDFW����IURP�WKH�ODVW�WHUP��7R�ILQG�WKH�next term, subtract 15 from the resulting number, and so on. � ±60 ± 15 = ±75 ±75 ± 15 = ±90 ±90 ± 15 = ±105 � So, the next three terms of this arithmetic sequence are ±75, ±90, ±105.

����5, 8, 11, 14, «

62/87,21���Find the common difference by subtracting two consecutive terms. � 8 ± �� ���� � The common difference between terms is 3. So, to find the next term, add 3 from the last term. To find the next term, add 3 from the resulting number, and so on. � 14 + 3 = 17 17 + 3 = 20 20 + 3 = 23 � So, the next three terms of this arithmetic sequence are 17, 20, 23.

Determine whether each sequence is an arithmetic sequence. If it is, state the common difference.����±40, ±32, ±24, ±16, «

62/87,21���±40, ±32, ±24, ±16, « � An arithmetic sequence is a numerical pattern that increases or decreases at a constant rate called the common difference. To find the common difference, subtract two consecutive numbers in the sequence. � ±32 ± (±40) = 8 ±24 ± (±32) = 8 ±16 ± (±24) = 7 � 7KH�GLIIHUHQFH�EHWZHHQ�WHUPV�LV�FRQVWDQW��VR�WKH�VHTXHQFH�LV�DQ�DULWKPHWLF�VHTXHQFH�� The common difference is 4. �

����0.75, 1.5, 3, 6, 12, «

62/87,21���An arithmetic sequence is a numerical pattern that increases or decreases at a constant rate called the common difference. To find the common difference, subtract two consecutive numbers in the sequence. � 1.5 ± 0.75 = 0.75 3 ± 1.5 = 1.5 6 ± 3 = 3 12 ± 6 = 6 � The difference between terms is not constant, so the sequence is not an an arithmetic sequence and does not have a common difference�� �

����5, 17, 29, 41, «

62/87,21���An arithmetic sequence is a numerical pattern that increases or decreases at a constant rate called the common difference. To find the common difference, subtract two consecutive numbers in the sequence. � 17 ± 5 = 12 29 ± 17 = 12 41 ± 29 = 12 � 7KH�GLIIHUHQFH�EHWZHHQ�WHUPV�LV�FRQVWDQW��VR�WKH�VHTXHQFH�LV�DQ�DULWKPHWLF�VHTXHQFH�� The common difference is 4. �

����MULTIPLE CHOICE In each figure, only one side of each regular pentagon is shared with another pentagon. Each side of each pentagon is 1 centimeter. If the pattern continues, what is the perimeter of a figure that has 6 pentagons?

F 15 cm H 20 cm G 25 cm J 30 cm

62/87,21���

� The perimeter increases by 3 as the number of polygons increases by 1. Continue the sequence to find the perimeter of a figure that has 6 pentagons. �

� So, a figure that has 6 pentagons has a perimeter of 20 centimeters. The correct choice is H.

Number of Pentagons

Perimeter

1 5 2 8 3 11

Number of Pentagons

Perimeter

4 14 5 17 6 20

eSolutions Manual - Powered by Cognero Page 10

Practice Test - Chapter 3

���TEMPERATURE The equation to convert Celsius temperature C to Kelvin temperature K is shown. �

� a. State the independent and dependent variables.Explain. b. Determine the C- and K-intercepts and describe what the intercepts mean in this situation.

62/87,21���a. The Kelvin temperature is dependent on the Celsius temperature. So, the Celsius temperature is the independent variable and the Kelvin temperature is the dependent variable. � b. The C-intercept is (±273, 0) and it means that a Celsius temperature of ±273 degrees is equal to a Kelvin temperature of 0 degrees. The K-intercept is (0, 273) and it means that a Celsius temperature of 0 degrees is equal to a Kelvin temperature of 273 degrees.

Graph each equation.���y = x + 2

62/87,21���To graph the equation, find the x- and y-intercepts. Plot these two points. Then draw a line through them.To find the x-intercept, let y = 0. �

� To find the y-intercept, let x = 0. �

� So, the x-intercept is ±2 and the y-intercept is 2. �

���y = 4x

62/87,21���The slope of y = 4x is 4. Graph (0, 0). From there, move up 4 units and right 1 unit to find another point. Then draw aline containing the points. �

���x + 2y = ±1

62/87,21���To graph the equation, find the x- and y-intercepts. Plot these two points. Then draw a line through them.To find the x-intercept, let y = 0. �

� To find the y-intercept, let x = 0. �

So, the x-intercept is ±1 and the y-intercept is .

�

���±3x = 5 ± y

62/87,21���To graph the equation, find the x- and y-intercepts. Plot these two points. Then draw a line through them.To find the x-intercept, let y = 0. �

� To find the y-intercept, let x = 0. �

�

So, the x-intercept is �DQG�WKH�y-intercept is 5.

�

�

Solve each equation by graphing.���4x + 2 = 0

62/87,21���The related function is y = 4x + 2. 7R�JUDSK�WKH�IXQFWLRQ��PDNH�D�WDEOH���

The graph intersects the x-axis at . So, the solution is .

x f (x ) = 4x + 2 f (x �� (x, f(x ) ) ±4 f (±4) = 4(±4) + 2 ±14 (±4, ±14) ±2 f (±2) = 4(±2) + 2 ±6 (±2, ±6) ±0.5 f(±0.5) = 4(±0.5) + 2 0 (±0.5, 0)

0 f (0) = 4(0) + 2 2 (0, 2) 2 f (2) = 4(2) + 2 10 (2, 10) 4 f (4) = 4(4) + 2 18 (4, 18)

���0 = 6 ± 3x

62/87,21���The related function is y = í3x + 6. 7R�JUDSK�WKH�IXQFWLRQ��PDNH�D�WDEOH���

�

The graph intersects the x-axis at 2. So, the solution is 2.

x f (x ) = ±3x + 6 f (x �� (x, f(x ) ) ±4 f (±4) = ±3(±4) + 6 18 (±4, 18) ±2 f (±2) = ±3(±2) + 6 12 (±2, 12) 0 f (0) = ±3(0) + 6 6 (0, 6) 2 f (2) = ±3(2) + 6 0 (2, 0) 3 f (3) = ±3(3) + 6 ±3 (3, ±3) 4 f (4) = ±3(4) + 6 ±6 (4, ±6)

���5x + 2 = ±3

62/87,21���The related function is y = 5x + 5. 7R�JUDSK�WKH�IXQFWLRQ��PDNH�D�WDEOH���

�

The graph intersects the x-axis at ±1. So, the solution is ±1.

x f (x ) = 5x + 5 f (x �� (x, f(x ) ) ±4 f (±4) = 5(±4) + 5 ±15 (±4, ±15) ±2 f (±2) = 5(±2) + 5 ±5 (±2, ±5) ±1 f (±1) = 5(±1) + 5 0 (±1, 0) 0 f (0) = 5(0) + 5 5 (0, 5) 2 f (2) = 5(2) + 5 15 (2, 15) 4 f (4) = 5(4) + 5 25 (4, 25)

���12x = 4x + 16

62/87,21���The related function is y = í8x��������7R�JUDSK�WKH�IXQFWLRQ��PDNH�D�WDEOH���

�

The graph intersects the x-axis at 2. So, the solution is 2.

x f (x ) = ±8x + 16 f (x �� (x, f(x ) ) ±4 f (±4) = ±8(±4) + 16 48 (±4, 48) ±2 f (±2) = ±8(±2) + 16 32 (±2, 32) 0 f (0) = ±8(0) + 16 16 (0, 16) 2 f (2) = ±8(2) + 16 0 (2, 0) 3 f (3) = ±8(3) + 16 ±8 (3, ±8) 4 f (4) = ±8(4) + 16 ±16 (4, ±16)

Find the slope of the line that passes through each pair of points.����(5, 8), (±3, 7)

62/87,21���

So, the slope is .

����(5, ±2), (3, ±2)

62/87,21���

� So, the slope is 0.

����(±4, 7), (8, ±1)

62/87,21���

So, the slope is .

����(6, ±3), (6, 4)

62/87,21���

� So, the slope is undefined.

����MULTIPLE CHOICE Which is the slope of the linear function shown in the graph?�

A

B �

C

D

62/87,21���The line passes through the points (±2, 3) and (3, 1).�

The slope is , so the correct choice is B.

Suppose y varies directly as x. Write a direct variation equation that relates x and y . Then solve.����If y = 6 when x = 9, find x when y = 12.

62/87,21���

� So, the direct variation equation is ����Substitute 12 for y and find x.´��

� So, x = 18 when y = 12.

����When y = ±8, x = 8. What is x when y = ±6?

62/87,21���

� So, the direct variation equation is y = ±x���Substitute ±6 for y and find x.´��

� So, x = 6 when y = ±6.

����If y = ±5 when x = ±2, what is y when x = 14?

62/87,21���

� So, the direct variation equation is ���Substitute 14 for x and find y .�

� So, y = 35 when x = 14.

����If y = 2 when x = ±12, find y when x = ±4.

62/87,21���

� So, the direct variation equation is ����Substitute ±4 for x and find y .´��

�

So, y = �ZKHQ�x = ±4.

����BIOLOGY The number of pints of blood in a human body varies directly with the person¶s weight. A person who weighs 120 pounds has about 8.4 pints of blood in his or her body. a. Write and graph an equation relating weight and amount of blood in a person¶s body. b. Predict the weight of a person whose body holds 12 pints of blood.

62/87,21���a. To write a direct variation equation, find the constant of variation k . Let x = 120 and y = 8.4.�

� So, the direct variation equation is y = 0.07x.

b. Using the direct variation equation from part a, let y = 12. �

� So, a person who holds 12 pints of blood would weigh about 171 pounds.

Find the next three terms of each arithmetic sequence.����0, ±15, ±30, ±45, ±60, «

62/87,21���Find the common difference by subtracting two consecutive terms. � ±15 ± 0 = ±���� � The common difference between terms is ±�����6R��WR�ILQG�WKH�QH[W�WHUP��VXEWUDFW����IURP�WKH�ODVW�WHUP��7R�ILQG�WKH�next term, subtract 15 from the resulting number, and so on. � ±60 ± 15 = ±75 ±75 ± 15 = ±90 ±90 ± 15 = ±105 � So, the next three terms of this arithmetic sequence are ±75, ±90, ±105.

����5, 8, 11, 14, «

62/87,21���Find the common difference by subtracting two consecutive terms. � 8 ± �� ���� � The common difference between terms is 3. So, to find the next term, add 3 from the last term. To find the next term, add 3 from the resulting number, and so on. � 14 + 3 = 17 17 + 3 = 20 20 + 3 = 23 � So, the next three terms of this arithmetic sequence are 17, 20, 23.

Determine whether each sequence is an arithmetic sequence. If it is, state the common difference.����±40, ±32, ±24, ±16, «

62/87,21���±40, ±32, ±24, ±16, « � An arithmetic sequence is a numerical pattern that increases or decreases at a constant rate called the common difference. To find the common difference, subtract two consecutive numbers in the sequence. � ±32 ± (±40) = 8 ±24 ± (±32) = 8 ±16 ± (±24) = 7 � 7KH�GLIIHUHQFH�EHWZHHQ�WHUPV�LV�FRQVWDQW��VR�WKH�VHTXHQFH�LV�DQ�DULWKPHWLF�VHTXHQFH�� The common difference is 4. �

����0.75, 1.5, 3, 6, 12, «

62/87,21���An arithmetic sequence is a numerical pattern that increases or decreases at a constant rate called the common difference. To find the common difference, subtract two consecutive numbers in the sequence. � 1.5 ± 0.75 = 0.75 3 ± 1.5 = 1.5 6 ± 3 = 3 12 ± 6 = 6 � The difference between terms is not constant, so the sequence is not an an arithmetic sequence and does not have a common difference�� �

����5, 17, 29, 41, «

62/87,21���An arithmetic sequence is a numerical pattern that increases or decreases at a constant rate called the common difference. To find the common difference, subtract two consecutive numbers in the sequence. � 17 ± 5 = 12 29 ± 17 = 12 41 ± 29 = 12 � 7KH�GLIIHUHQFH�EHWZHHQ�WHUPV�LV�FRQVWDQW��VR�WKH�VHTXHQFH�LV�DQ�DULWKPHWLF�VHTXHQFH�� The common difference is 4. �

����MULTIPLE CHOICE In each figure, only one side of each regular pentagon is shared with another pentagon. Each side of each pentagon is 1 centimeter. If the pattern continues, what is the perimeter of a figure that has 6 pentagons?

F 15 cm H 20 cm G 25 cm J 30 cm

62/87,21���

� The perimeter increases by 3 as the number of polygons increases by 1. Continue the sequence to find the perimeter of a figure that has 6 pentagons. �

� So, a figure that has 6 pentagons has a perimeter of 20 centimeters. The correct choice is H.

Number of Pentagons

Perimeter

1 5 2 8 3 11

Number of Pentagons

Perimeter

4 14 5 17 6 20

eSolutions Manual - Powered by Cognero Page 11

Practice Test - Chapter 3

���TEMPERATURE The equation to convert Celsius temperature C to Kelvin temperature K is shown. �

� a. State the independent and dependent variables.Explain. b. Determine the C- and K-intercepts and describe what the intercepts mean in this situation.

62/87,21���a. The Kelvin temperature is dependent on the Celsius temperature. So, the Celsius temperature is the independent variable and the Kelvin temperature is the dependent variable. � b. The C-intercept is (±273, 0) and it means that a Celsius temperature of ±273 degrees is equal to a Kelvin temperature of 0 degrees. The K-intercept is (0, 273) and it means that a Celsius temperature of 0 degrees is equal to a Kelvin temperature of 273 degrees.

Graph each equation.���y = x + 2

62/87,21���To graph the equation, find the x- and y-intercepts. Plot these two points. Then draw a line through them.To find the x-intercept, let y = 0. �

� To find the y-intercept, let x = 0. �

� So, the x-intercept is ±2 and the y-intercept is 2. �

���y = 4x

62/87,21���The slope of y = 4x is 4. Graph (0, 0). From there, move up 4 units and right 1 unit to find another point. Then draw aline containing the points. �

���x + 2y = ±1

62/87,21���To graph the equation, find the x- and y-intercepts. Plot these two points. Then draw a line through them.To find the x-intercept, let y = 0. �

� To find the y-intercept, let x = 0. �

So, the x-intercept is ±1 and the y-intercept is .

�

���±3x = 5 ± y

62/87,21���To graph the equation, find the x- and y-intercepts. Plot these two points. Then draw a line through them.To find the x-intercept, let y = 0. �

� To find the y-intercept, let x = 0. �

�

So, the x-intercept is �DQG�WKH�y-intercept is 5.

�

�

Solve each equation by graphing.���4x + 2 = 0

62/87,21���The related function is y = 4x + 2. 7R�JUDSK�WKH�IXQFWLRQ��PDNH�D�WDEOH���

The graph intersects the x-axis at . So, the solution is .

x f (x ) = 4x + 2 f (x �� (x, f(x ) ) ±4 f (±4) = 4(±4) + 2 ±14 (±4, ±14) ±2 f (±2) = 4(±2) + 2 ±6 (±2, ±6) ±0.5 f(±0.5) = 4(±0.5) + 2 0 (±0.5, 0)

0 f (0) = 4(0) + 2 2 (0, 2) 2 f (2) = 4(2) + 2 10 (2, 10) 4 f (4) = 4(4) + 2 18 (4, 18)

���0 = 6 ± 3x

62/87,21���The related function is y = í3x + 6. 7R�JUDSK�WKH�IXQFWLRQ��PDNH�D�WDEOH���

�

The graph intersects the x-axis at 2. So, the solution is 2.

x f (x ) = ±3x + 6 f (x �� (x, f(x ) ) ±4 f (±4) = ±3(±4) + 6 18 (±4, 18) ±2 f (±2) = ±3(±2) + 6 12 (±2, 12) 0 f (0) = ±3(0) + 6 6 (0, 6) 2 f (2) = ±3(2) + 6 0 (2, 0) 3 f (3) = ±3(3) + 6 ±3 (3, ±3) 4 f (4) = ±3(4) + 6 ±6 (4, ±6)

���5x + 2 = ±3

62/87,21���The related function is y = 5x + 5. 7R�JUDSK�WKH�IXQFWLRQ��PDNH�D�WDEOH���

�

The graph intersects the x-axis at ±1. So, the solution is ±1.

x f (x ) = 5x + 5 f (x �� (x, f(x ) ) ±4 f (±4) = 5(±4) + 5 ±15 (±4, ±15) ±2 f (±2) = 5(±2) + 5 ±5 (±2, ±5) ±1 f (±1) = 5(±1) + 5 0 (±1, 0) 0 f (0) = 5(0) + 5 5 (0, 5) 2 f (2) = 5(2) + 5 15 (2, 15) 4 f (4) = 5(4) + 5 25 (4, 25)

���12x = 4x + 16

62/87,21���The related function is y = í8x��������7R�JUDSK�WKH�IXQFWLRQ��PDNH�D�WDEOH���

�

The graph intersects the x-axis at 2. So, the solution is 2.

x f (x ) = ±8x + 16 f (x �� (x, f(x ) ) ±4 f (±4) = ±8(±4) + 16 48 (±4, 48) ±2 f (±2) = ±8(±2) + 16 32 (±2, 32) 0 f (0) = ±8(0) + 16 16 (0, 16) 2 f (2) = ±8(2) + 16 0 (2, 0) 3 f (3) = ±8(3) + 16 ±8 (3, ±8) 4 f (4) = ±8(4) + 16 ±16 (4, ±16)

Find the slope of the line that passes through each pair of points.����(5, 8), (±3, 7)

62/87,21���

So, the slope is .

����(5, ±2), (3, ±2)

62/87,21���

� So, the slope is 0.

����(±4, 7), (8, ±1)

62/87,21���

So, the slope is .

����(6, ±3), (6, 4)

62/87,21���

� So, the slope is undefined.

����MULTIPLE CHOICE Which is the slope of the linear function shown in the graph?�

A

B �

C

D

62/87,21���The line passes through the points (±2, 3) and (3, 1).�

The slope is , so the correct choice is B.

Suppose y varies directly as x. Write a direct variation equation that relates x and y . Then solve.����If y = 6 when x = 9, find x when y = 12.

62/87,21���

� So, the direct variation equation is ����Substitute 12 for y and find x.´��

� So, x = 18 when y = 12.

����When y = ±8, x = 8. What is x when y = ±6?

62/87,21���

� So, the direct variation equation is y = ±x���Substitute ±6 for y and find x.´��

� So, x = 6 when y = ±6.

����If y = ±5 when x = ±2, what is y when x = 14?

62/87,21���

� So, the direct variation equation is ���Substitute 14 for x and find y .�

� So, y = 35 when x = 14.

����If y = 2 when x = ±12, find y when x = ±4.

62/87,21���

� So, the direct variation equation is ����Substitute ±4 for x and find y .´��

�

So, y = �ZKHQ�x = ±4.

����BIOLOGY The number of pints of blood in a human body varies directly with the person¶s weight. A person who weighs 120 pounds has about 8.4 pints of blood in his or her body. a. Write and graph an equation relating weight and amount of blood in a person¶s body. b. Predict the weight of a person whose body holds 12 pints of blood.

62/87,21���a. To write a direct variation equation, find the constant of variation k . Let x = 120 and y = 8.4.�

� So, the direct variation equation is y = 0.07x.

b. Using the direct variation equation from part a, let y = 12. �

� So, a person who holds 12 pints of blood would weigh about 171 pounds.

Find the next three terms of each arithmetic sequence.����0, ±15, ±30, ±45, ±60, «

62/87,21���Find the common difference by subtracting two consecutive terms. � ±15 ± 0 = ±���� � The common difference between terms is ±�����6R��WR�ILQG�WKH�QH[W�WHUP��VXEWUDFW����IURP�WKH�ODVW�WHUP��7R�ILQG�WKH�next term, subtract 15 from the resulting number, and so on. � ±60 ± 15 = ±75 ±75 ± 15 = ±90 ±90 ± 15 = ±105 � So, the next three terms of this arithmetic sequence are ±75, ±90, ±105.

����5, 8, 11, 14, «

62/87,21���Find the common difference by subtracting two consecutive terms. � 8 ± �� ���� � The common difference between terms is 3. So, to find the next term, add 3 from the last term. To find the next term, add 3 from the resulting number, and so on. � 14 + 3 = 17 17 + 3 = 20 20 + 3 = 23 � So, the next three terms of this arithmetic sequence are 17, 20, 23.

Determine whether each sequence is an arithmetic sequence. If it is, state the common difference.����±40, ±32, ±24, ±16, «

62/87,21���±40, ±32, ±24, ±16, « � An arithmetic sequence is a numerical pattern that increases or decreases at a constant rate called the common difference. To find the common difference, subtract two consecutive numbers in the sequence. � ±32 ± (±40) = 8 ±24 ± (±32) = 8 ±16 ± (±24) = 7 � 7KH�GLIIHUHQFH�EHWZHHQ�WHUPV�LV�FRQVWDQW��VR�WKH�VHTXHQFH�LV�DQ�DULWKPHWLF�VHTXHQFH�� The common difference is 4. �

����0.75, 1.5, 3, 6, 12, «

62/87,21���An arithmetic sequence is a numerical pattern that increases or decreases at a constant rate called the common difference. To find the common difference, subtract two consecutive numbers in the sequence. � 1.5 ± 0.75 = 0.75 3 ± 1.5 = 1.5 6 ± 3 = 3 12 ± 6 = 6 � The difference between terms is not constant, so the sequence is not an an arithmetic sequence and does not have a common difference�� �

����5, 17, 29, 41, «

62/87,21���An arithmetic sequence is a numerical pattern that increases or decreases at a constant rate called the common difference. To find the common difference, subtract two consecutive numbers in the sequence. � 17 ± 5 = 12 29 ± 17 = 12 41 ± 29 = 12 � 7KH�GLIIHUHQFH�EHWZHHQ�WHUPV�LV�FRQVWDQW��VR�WKH�VHTXHQFH�LV�DQ�DULWKPHWLF�VHTXHQFH�� The common difference is 4. �

����MULTIPLE CHOICE In each figure, only one side of each regular pentagon is shared with another pentagon. Each side of each pentagon is 1 centimeter. If the pattern continues, what is the perimeter of a figure that has 6 pentagons?

F 15 cm H 20 cm G 25 cm J 30 cm

62/87,21���

� The perimeter increases by 3 as the number of polygons increases by 1. Continue the sequence to find the perimeter of a figure that has 6 pentagons. �

� So, a figure that has 6 pentagons has a perimeter of 20 centimeters. The correct choice is H.

Number of Pentagons

Perimeter

1 5 2 8 3 11

Number of Pentagons

Perimeter

4 14 5 17 6 20

eSolutions Manual - Powered by Cognero Page 12

Practice Test - Chapter 3

���TEMPERATURE The equation to convert Celsius temperature C to Kelvin temperature K is shown. �

� a. State the independent and dependent variables.Explain. b. Determine the C- and K-intercepts and describe what the intercepts mean in this situation.

62/87,21���a. The Kelvin temperature is dependent on the Celsius temperature. So, the Celsius temperature is the independent variable and the Kelvin temperature is the dependent variable. � b. The C-intercept is (±273, 0) and it means that a Celsius temperature of ±273 degrees is equal to a Kelvin temperature of 0 degrees. The K-intercept is (0, 273) and it means that a Celsius temperature of 0 degrees is equal to a Kelvin temperature of 273 degrees.

Graph each equation.���y = x + 2

62/87,21���To graph the equation, find the x- and y-intercepts. Plot these two points. Then draw a line through them.To find the x-intercept, let y = 0. �

� To find the y-intercept, let x = 0. �

� So, the x-intercept is ±2 and the y-intercept is 2. �

���y = 4x

62/87,21���The slope of y = 4x is 4. Graph (0, 0). From there, move up 4 units and right 1 unit to find another point. Then draw aline containing the points. �

���x + 2y = ±1

62/87,21���To graph the equation, find the x- and y-intercepts. Plot these two points. Then draw a line through them.To find the x-intercept, let y = 0. �

� To find the y-intercept, let x = 0. �

So, the x-intercept is ±1 and the y-intercept is .

�

���±3x = 5 ± y

62/87,21���To graph the equation, find the x- and y-intercepts. Plot these two points. Then draw a line through them.To find the x-intercept, let y = 0. �

� To find the y-intercept, let x = 0. �

�

So, the x-intercept is �DQG�WKH�y-intercept is 5.

�

�

Solve each equation by graphing.���4x + 2 = 0

62/87,21���The related function is y = 4x + 2. 7R�JUDSK�WKH�IXQFWLRQ��PDNH�D�WDEOH���

The graph intersects the x-axis at . So, the solution is .

x f (x ) = 4x + 2 f (x �� (x, f(x ) ) ±4 f (±4) = 4(±4) + 2 ±14 (±4, ±14) ±2 f (±2) = 4(±2) + 2 ±6 (±2, ±6) ±0.5 f(±0.5) = 4(±0.5) + 2 0 (±0.5, 0)

0 f (0) = 4(0) + 2 2 (0, 2) 2 f (2) = 4(2) + 2 10 (2, 10) 4 f (4) = 4(4) + 2 18 (4, 18)

���0 = 6 ± 3x

62/87,21���The related function is y = í3x + 6. 7R�JUDSK�WKH�IXQFWLRQ��PDNH�D�WDEOH���

�

The graph intersects the x-axis at 2. So, the solution is 2.

x f (x ) = ±3x + 6 f (x �� (x, f(x ) ) ±4 f (±4) = ±3(±4) + 6 18 (±4, 18) ±2 f (±2) = ±3(±2) + 6 12 (±2, 12) 0 f (0) = ±3(0) + 6 6 (0, 6) 2 f (2) = ±3(2) + 6 0 (2, 0) 3 f (3) = ±3(3) + 6 ±3 (3, ±3) 4 f (4) = ±3(4) + 6 ±6 (4, ±6)

���5x + 2 = ±3

62/87,21���The related function is y = 5x + 5. 7R�JUDSK�WKH�IXQFWLRQ��PDNH�D�WDEOH���

�

The graph intersects the x-axis at ±1. So, the solution is ±1.

x f (x ) = 5x + 5 f (x �� (x, f(x ) ) ±4 f (±4) = 5(±4) + 5 ±15 (±4, ±15) ±2 f (±2) = 5(±2) + 5 ±5 (±2, ±5) ±1 f (±1) = 5(±1) + 5 0 (±1, 0) 0 f (0) = 5(0) + 5 5 (0, 5) 2 f (2) = 5(2) + 5 15 (2, 15) 4 f (4) = 5(4) + 5 25 (4, 25)

���12x = 4x + 16

62/87,21���The related function is y = í8x��������7R�JUDSK�WKH�IXQFWLRQ��PDNH�D�WDEOH���

�

The graph intersects the x-axis at 2. So, the solution is 2.

x f (x ) = ±8x + 16 f (x �� (x, f(x ) ) ±4 f (±4) = ±8(±4) + 16 48 (±4, 48) ±2 f (±2) = ±8(±2) + 16 32 (±2, 32) 0 f (0) = ±8(0) + 16 16 (0, 16) 2 f (2) = ±8(2) + 16 0 (2, 0) 3 f (3) = ±8(3) + 16 ±8 (3, ±8) 4 f (4) = ±8(4) + 16 ±16 (4, ±16)

Find the slope of the line that passes through each pair of points.����(5, 8), (±3, 7)

62/87,21���

So, the slope is .

����(5, ±2), (3, ±2)

62/87,21���

� So, the slope is 0.

����(±4, 7), (8, ±1)

62/87,21���

So, the slope is .

����(6, ±3), (6, 4)

62/87,21���

� So, the slope is undefined.

����MULTIPLE CHOICE Which is the slope of the linear function shown in the graph?�

A

B �

C

D

62/87,21���The line passes through the points (±2, 3) and (3, 1).�

The slope is , so the correct choice is B.

Suppose y varies directly as x. Write a direct variation equation that relates x and y . Then solve.����If y = 6 when x = 9, find x when y = 12.

62/87,21���

� So, the direct variation equation is ����Substitute 12 for y and find x.´��

� So, x = 18 when y = 12.

����When y = ±8, x = 8. What is x when y = ±6?

62/87,21���

� So, the direct variation equation is y = ±x���Substitute ±6 for y and find x.´��

� So, x = 6 when y = ±6.

����If y = ±5 when x = ±2, what is y when x = 14?

62/87,21���

� So, the direct variation equation is ���Substitute 14 for x and find y .�

� So, y = 35 when x = 14.

����If y = 2 when x = ±12, find y when x = ±4.

62/87,21���

� So, the direct variation equation is ����Substitute ±4 for x and find y .´��

�

So, y = �ZKHQ�x = ±4.

����BIOLOGY The number of pints of blood in a human body varies directly with the person¶s weight. A person who weighs 120 pounds has about 8.4 pints of blood in his or her body. a. Write and graph an equation relating weight and amount of blood in a person¶s body. b. Predict the weight of a person whose body holds 12 pints of blood.

62/87,21���a. To write a direct variation equation, find the constant of variation k . Let x = 120 and y = 8.4.�

� So, the direct variation equation is y = 0.07x.

b. Using the direct variation equation from part a, let y = 12. �

� So, a person who holds 12 pints of blood would weigh about 171 pounds.

Find the next three terms of each arithmetic sequence.����0, ±15, ±30, ±45, ±60, «

62/87,21���Find the common difference by subtracting two consecutive terms. � ±15 ± 0 = ±���� � The common difference between terms is ±�����6R��WR�ILQG�WKH�QH[W�WHUP��VXEWUDFW����IURP�WKH�ODVW�WHUP��7R�ILQG�WKH�next term, subtract 15 from the resulting number, and so on. � ±60 ± 15 = ±75 ±75 ± 15 = ±90 ±90 ± 15 = ±105 � So, the next three terms of this arithmetic sequence are ±75, ±90, ±105.

����5, 8, 11, 14, «

62/87,21���Find the common difference by subtracting two consecutive terms. � 8 ± �� ���� � The common difference between terms is 3. So, to find the next term, add 3 from the last term. To find the next term, add 3 from the resulting number, and so on. � 14 + 3 = 17 17 + 3 = 20 20 + 3 = 23 � So, the next three terms of this arithmetic sequence are 17, 20, 23.

Determine whether each sequence is an arithmetic sequence. If it is, state the common difference.����±40, ±32, ±24, ±16, «

62/87,21���±40, ±32, ±24, ±16, « � An arithmetic sequence is a numerical pattern that increases or decreases at a constant rate called the common difference. To find the common difference, subtract two consecutive numbers in the sequence. � ±32 ± (±40) = 8 ±24 ± (±32) = 8 ±16 ± (±24) = 7 � 7KH�GLIIHUHQFH�EHWZHHQ�WHUPV�LV�FRQVWDQW��VR�WKH�VHTXHQFH�LV�DQ�DULWKPHWLF�VHTXHQFH�� The common difference is 4. �

����0.75, 1.5, 3, 6, 12, «

62/87,21���An arithmetic sequence is a numerical pattern that increases or decreases at a constant rate called the common difference. To find the common difference, subtract two consecutive numbers in the sequence. � 1.5 ± 0.75 = 0.75 3 ± 1.5 = 1.5 6 ± 3 = 3 12 ± 6 = 6 � The difference between terms is not constant, so the sequence is not an an arithmetic sequence and does not have a common difference�� �

����5, 17, 29, 41, «

62/87,21���An arithmetic sequence is a numerical pattern that increases or decreases at a constant rate called the common difference. To find the common difference, subtract two consecutive numbers in the sequence. � 17 ± 5 = 12 29 ± 17 = 12 41 ± 29 = 12 � 7KH�GLIIHUHQFH�EHWZHHQ�WHUPV�LV�FRQVWDQW��VR�WKH�VHTXHQFH�LV�DQ�DULWKPHWLF�VHTXHQFH�� The common difference is 4. �

����MULTIPLE CHOICE In each figure, only one side of each regular pentagon is shared with another pentagon. Each side of each pentagon is 1 centimeter. If the pattern continues, what is the perimeter of a figure that has 6 pentagons?

F 15 cm H 20 cm G 25 cm J 30 cm

62/87,21���

� The perimeter increases by 3 as the number of polygons increases by 1. Continue the sequence to find the perimeter of a figure that has 6 pentagons. �

� So, a figure that has 6 pentagons has a perimeter of 20 centimeters. The correct choice is H.

Number of Pentagons

Perimeter

1 5 2 8 3 11

Number of Pentagons

Perimeter

4 14 5 17 6 20

eSolutions Manual - Powered by Cognero Page 13

Practice Test - Chapter 3

���TEMPERATURE The equation to convert Celsius temperature C to Kelvin temperature K is shown. �

� a. State the independent and dependent variables.Explain. b. Determine the C- and K-intercepts and describe what the intercepts mean in this situation.

62/87,21���a. The Kelvin temperature is dependent on the Celsius temperature. So, the Celsius temperature is the independent variable and the Kelvin temperature is the dependent variable. � b. The C-intercept is (±273, 0) and it means that a Celsius temperature of ±273 degrees is equal to a Kelvin temperature of 0 degrees. The K-intercept is (0, 273) and it means that a Celsius temperature of 0 degrees is equal to a Kelvin temperature of 273 degrees.

Graph each equation.���y = x + 2

62/87,21���To graph the equation, find the x- and y-intercepts. Plot these two points. Then draw a line through them.To find the x-intercept, let y = 0. �

� To find the y-intercept, let x = 0. �

� So, the x-intercept is ±2 and the y-intercept is 2. �

���y = 4x

62/87,21���The slope of y = 4x is 4. Graph (0, 0). From there, move up 4 units and right 1 unit to find another point. Then draw aline containing the points. �

���x + 2y = ±1

62/87,21���To graph the equation, find the x- and y-intercepts. Plot these two points. Then draw a line through them.To find the x-intercept, let y = 0. �

� To find the y-intercept, let x = 0. �

So, the x-intercept is ±1 and the y-intercept is .

�

���±3x = 5 ± y

62/87,21���To graph the equation, find the x- and y-intercepts. Plot these two points. Then draw a line through them.To find the x-intercept, let y = 0. �

� To find the y-intercept, let x = 0. �

�

So, the x-intercept is �DQG�WKH�y-intercept is 5.

�

�

Solve each equation by graphing.���4x + 2 = 0

62/87,21���The related function is y = 4x + 2. 7R�JUDSK�WKH�IXQFWLRQ��PDNH�D�WDEOH���

The graph intersects the x-axis at . So, the solution is .

x f (x ) = 4x + 2 f (x �� (x, f(x ) ) ±4 f (±4) = 4(±4) + 2 ±14 (±4, ±14) ±2 f (±2) = 4(±2) + 2 ±6 (±2, ±6) ±0.5 f(±0.5) = 4(±0.5) + 2 0 (±0.5, 0)

0 f (0) = 4(0) + 2 2 (0, 2) 2 f (2) = 4(2) + 2 10 (2, 10) 4 f (4) = 4(4) + 2 18 (4, 18)

���0 = 6 ± 3x

62/87,21���The related function is y = í3x + 6. 7R�JUDSK�WKH�IXQFWLRQ��PDNH�D�WDEOH���

�

The graph intersects the x-axis at 2. So, the solution is 2.

x f (x ) = ±3x + 6 f (x �� (x, f(x ) ) ±4 f (±4) = ±3(±4) + 6 18 (±4, 18) ±2 f (±2) = ±3(±2) + 6 12 (±2, 12) 0 f (0) = ±3(0) + 6 6 (0, 6) 2 f (2) = ±3(2) + 6 0 (2, 0) 3 f (3) = ±3(3) + 6 ±3 (3, ±3) 4 f (4) = ±3(4) + 6 ±6 (4, ±6)

���5x + 2 = ±3

62/87,21���The related function is y = 5x + 5. 7R�JUDSK�WKH�IXQFWLRQ��PDNH�D�WDEOH���

�

The graph intersects the x-axis at ±1. So, the solution is ±1.

x f (x ) = 5x + 5 f (x �� (x, f(x ) ) ±4 f (±4) = 5(±4) + 5 ±15 (±4, ±15) ±2 f (±2) = 5(±2) + 5 ±5 (±2, ±5) ±1 f (±1) = 5(±1) + 5 0 (±1, 0) 0 f (0) = 5(0) + 5 5 (0, 5) 2 f (2) = 5(2) + 5 15 (2, 15) 4 f (4) = 5(4) + 5 25 (4, 25)

���12x = 4x + 16

62/87,21���The related function is y = í8x��������7R�JUDSK�WKH�IXQFWLRQ��PDNH�D�WDEOH���

�

The graph intersects the x-axis at 2. So, the solution is 2.

x f (x ) = ±8x + 16 f (x �� (x, f(x ) ) ±4 f (±4) = ±8(±4) + 16 48 (±4, 48) ±2 f (±2) = ±8(±2) + 16 32 (±2, 32) 0 f (0) = ±8(0) + 16 16 (0, 16) 2 f (2) = ±8(2) + 16 0 (2, 0) 3 f (3) = ±8(3) + 16 ±8 (3, ±8) 4 f (4) = ±8(4) + 16 ±16 (4, ±16)

Find the slope of the line that passes through each pair of points.����(5, 8), (±3, 7)

62/87,21���

So, the slope is .

����(5, ±2), (3, ±2)

62/87,21���

� So, the slope is 0.

����(±4, 7), (8, ±1)

62/87,21���

So, the slope is .

����(6, ±3), (6, 4)

62/87,21���

� So, the slope is undefined.

����MULTIPLE CHOICE Which is the slope of the linear function shown in the graph?�

A

B �

C

D

62/87,21���The line passes through the points (±2, 3) and (3, 1).�

The slope is , so the correct choice is B.

Suppose y varies directly as x. Write a direct variation equation that relates x and y . Then solve.����If y = 6 when x = 9, find x when y = 12.

62/87,21���

� So, the direct variation equation is ����Substitute 12 for y and find x.´��

� So, x = 18 when y = 12.

����When y = ±8, x = 8. What is x when y = ±6?

62/87,21���

� So, the direct variation equation is y = ±x���Substitute ±6 for y and find x.´��

� So, x = 6 when y = ±6.

����If y = ±5 when x = ±2, what is y when x = 14?

62/87,21���

� So, the direct variation equation is ���Substitute 14 for x and find y .�

� So, y = 35 when x = 14.

����If y = 2 when x = ±12, find y when x = ±4.

62/87,21���

� So, the direct variation equation is ����Substitute ±4 for x and find y .´��

�

So, y = �ZKHQ�x = ±4.

����BIOLOGY The number of pints of blood in a human body varies directly with the person¶s weight. A person who weighs 120 pounds has about 8.4 pints of blood in his or her body. a. Write and graph an equation relating weight and amount of blood in a person¶s body. b. Predict the weight of a person whose body holds 12 pints of blood.

62/87,21���a. To write a direct variation equation, find the constant of variation k . Let x = 120 and y = 8.4.�

� So, the direct variation equation is y = 0.07x.

b. Using the direct variation equation from part a, let y = 12. �

� So, a person who holds 12 pints of blood would weigh about 171 pounds.

Find the next three terms of each arithmetic sequence.����0, ±15, ±30, ±45, ±60, «

62/87,21���Find the common difference by subtracting two consecutive terms. � ±15 ± 0 = ±���� � The common difference between terms is ±�����6R��WR�ILQG�WKH�QH[W�WHUP��VXEWUDFW����IURP�WKH�ODVW�WHUP��7R�ILQG�WKH�next term, subtract 15 from the resulting number, and so on. � ±60 ± 15 = ±75 ±75 ± 15 = ±90 ±90 ± 15 = ±105 � So, the next three terms of this arithmetic sequence are ±75, ±90, ±105.

����5, 8, 11, 14, «

62/87,21���Find the common difference by subtracting two consecutive terms. � 8 ± �� ���� � The common difference between terms is 3. So, to find the next term, add 3 from the last term. To find the next term, add 3 from the resulting number, and so on. � 14 + 3 = 17 17 + 3 = 20 20 + 3 = 23 � So, the next three terms of this arithmetic sequence are 17, 20, 23.

Determine whether each sequence is an arithmetic sequence. If it is, state the common difference.����±40, ±32, ±24, ±16, «

62/87,21���±40, ±32, ±24, ±16, « � An arithmetic sequence is a numerical pattern that increases or decreases at a constant rate called the common difference. To find the common difference, subtract two consecutive numbers in the sequence. � ±32 ± (±40) = 8 ±24 ± (±32) = 8 ±16 ± (±24) = 7 � 7KH�GLIIHUHQFH�EHWZHHQ�WHUPV�LV�FRQVWDQW��VR�WKH�VHTXHQFH�LV�DQ�DULWKPHWLF�VHTXHQFH�� The common difference is 4. �

����0.75, 1.5, 3, 6, 12, «

62/87,21���An arithmetic sequence is a numerical pattern that increases or decreases at a constant rate called the common difference. To find the common difference, subtract two consecutive numbers in the sequence. � 1.5 ± 0.75 = 0.75 3 ± 1.5 = 1.5 6 ± 3 = 3 12 ± 6 = 6 � The difference between terms is not constant, so the sequence is not an an arithmetic sequence and does not have a common difference�� �

����5, 17, 29, 41, «

62/87,21���An arithmetic sequence is a numerical pattern that increases or decreases at a constant rate called the common difference. To find the common difference, subtract two consecutive numbers in the sequence. � 17 ± 5 = 12 29 ± 17 = 12 41 ± 29 = 12 � 7KH�GLIIHUHQFH�EHWZHHQ�WHUPV�LV�FRQVWDQW��VR�WKH�VHTXHQFH�LV�DQ�DULWKPHWLF�VHTXHQFH�� The common difference is 4. �

����MULTIPLE CHOICE In each figure, only one side of each regular pentagon is shared with another pentagon. Each side of each pentagon is 1 centimeter. If the pattern continues, what is the perimeter of a figure that has 6 pentagons?

F 15 cm H 20 cm G 25 cm J 30 cm

62/87,21���

� The perimeter increases by 3 as the number of polygons increases by 1. Continue the sequence to find the perimeter of a figure that has 6 pentagons. �

� So, a figure that has 6 pentagons has a perimeter of 20 centimeters. The correct choice is H.

Number of Pentagons

Perimeter

1 5 2 8 3 11

Number of Pentagons

Perimeter

4 14 5 17 6 20

eSolutions Manual - Powered by Cognero Page 14

Practice Test - Chapter 3