Download 1. TEMPERATURE The equation to convert Celsius temperature C

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.
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
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.
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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.
Practice
Test - Chapter
3 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
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
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 a
line containing the points.
x + 2y = 1
To graph the equation, find the x- and y-intercepts. Plot these two points. Then draw a line through them.
0.
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To find
the- x-intercept,
let y =
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Practice Test - Chapter 3
x + 2y = 1
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
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.
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Practice Test - Chapter 3
3x = 5 y
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
y-intercept is 5.
Solve each equation by graphing.
4x + 2 = 0
The related function is y = 4x + 2.
x
f (x ) = 4x + 2
f (x
4
f ( 4) = 4( 4) + 2
14
2
f ( 2) = 4( 2) + 2
6
0.5 f(- Powered
0.5) = 4(by0.5)
+2
0
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0
f (0) = 4(0) + 2
2
2
f (2) = 4(2) + 2
10
(x, f(x ) )
( 4, 14)
( 2, 6)
( 0.5, 0)
(0, 2)
(2, 10)
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Practice Test - Chapter 3
Solve each equation by graphing.
4x + 2 = 0
The related function is y = 4x + 2.
x
f (x ) = 4x + 2
f (x
f ( 4) = 4( 4) + 2
14
f ( 2) = 4( 2) + 2
6
f( 0.5) = 4( 0.5) + 2
0
f (0) = 4(0) + 2
2
f (2) = 4(2) + 2
10
f (4) = 4(4) + 2
18
4
2
0.5
0
2
4
(x, f(x ) )
( 4, 14)
( 2, 6)
( 0.5, 0)
(0, 2)
(2, 10)
(4, 18)
The graph intersects the x-axis at
0=6
. So, the solution is
.
3x
The related function is y = 3x + 6.
x
f (x ) = 3x + 6
4 f ( 4) = 3( 4) + 6
2 f ( 2) = 3( 2) + 6
0
f (0) = 3(0) + 6
2
f (2) = 3(2) + 6
3
f (3) = 3(3) + 6
4
f (4) = 3(4) + 6
f (x
18
12
6
0
3
6
(x, f(x ) )
( 4, 18)
( 2, 12)
(0, 6)
(2, 0)
(3, 3)
(4, 6)
The graph intersects the x-axis at 2. So, the solution is 2.
5x + 2 = 3
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The related function is y = 5x + 5.
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Practice Test - Chapter 3
The graph intersects the x-axis at 2. So, the solution is 2.
5x + 2 = 3
The related function is y = 5x + 5.
x
4
2
1
0
2
4
f (x ) = 5x + 5
f ( 4) = 5( 4) + 5
f ( 2) = 5( 2) + 5
f ( 1) = 5( 1) + 5
f (0) = 5(0) + 5
f (2) = 5(2) + 5
f (4) = 5(4) + 5
f (x
15
5
0
5
15
25
(x, f(x ) )
( 4, 15)
( 2, 5)
( 1, 0)
(0, 5)
(2, 15)
(4, 25)
The graph intersects the x-axis at 1. So, the solution is 1.
12x = 4x + 16
The related function is y = 8x
x
f (x ) = 8x + 16
4 f ( 4) = 8( 4) + 16
2 f ( 2) = 8( 2) + 16
0 f (0) = 8(0) + 16
2 f (2) = 8(2) + 16
3 f (3) = 8(3) + 16
4 f (4) = 8(4) + 16
f (x
48
32
16
0
8
16
(x, f(x ) )
( 4, 48)
( 2, 32)
(0, 16)
(2, 0)
(3, 8)
(4, 16)
The graph intersects the x-axis at 2. So, the solution is 2.
Find the slope of the line that passes through each pair of points.
(5, 8), ( 3, 7)
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Practice Test - Chapter 3
The graph intersects the x-axis at 2. So, the solution is 2.
Find the slope of the line that passes through each pair of points.
(5, 8), ( 3, 7)
So, the slope is
.
(5, 2), (3, 2)
So, the slope is 0.
( 4, 7), (8, 1)
So, the slope is
.
(6, 3), (6, 4)
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Practice
Test
- Chapter
So, the
slope
is
. 3
(6, 3), (6, 4)
So, the slope is undefined.
MULTIPLE CHOICE Which is the slope of the linear function shown in the graph?
A
B
C
D
The line passes through the points ( 2, 3) and (3, 1).
The slope is
, so the correct choice is B.
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Suppose
varies directly
x. Write a direct variation equation that relates x and y. Then solve.
If y = 6 when x = 9, find x when y = 12.
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Practice
Test is- Chapter
3 correct choice is B.
The slope
, so the
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.
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?
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?
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Practice Test - Chapter 3
So, x = 6 when y = 6.
If y = 5 when x = 2, what is y when x = 14?
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.
So, the direct variation equation is
So, y =
Substitute 4 for x and find y.
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.
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a. To write a direct variation equation, find the constant of variation k. Let x = 120 and y = 8.4.
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Practice
So, y Test
= - Chapter
x = 34.
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.
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,
Find the common difference by subtracting two consecutive terms.
15
0=
The common difference between terms is
next term, subtract 15 from the resulting number, and so on.
60
75
90
15 = 75
15 = 90
15 = 105
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So, the next three terms of this arithmetic sequence are 75, 90, 105.
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Practice Test - Chapter 3
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,
Find the common difference by subtracting two consecutive terms.
15
0=
The common difference between terms is
next term, subtract 15 from the resulting number, and so on.
60
75
90
15 = 75
15 = 90
15 = 105
So, the next three terms of this arithmetic sequence are 75, 90, 105.
5, 8, 11, 14,
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,
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
24
16
( 40) = 8
( 32) = 8
( 24) = 7
The common difference is 4.
0.75, 1.5, 3, 6, 12,
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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.
The common difference is 4.
Practice Test - Chapter 3
0.75, 1.5, 3, 6, 12,
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,
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
29
41
5 = 12
17 = 12
29 = 12
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
Number of
Pentagons
1
2
3
Perimeter
5
8
11
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.
Number
Perimeter
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Pentagons
4
14
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41
29 = 12
The common
difference
Practice
Test - Chapter
3 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
Number of
Pentagons
1
2
3
Perimeter
5
8
11
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.
Number of
Pentagons
4
5
6
Perimeter
14
17
20
So, a figure that has 6 pentagons has a perimeter of 20 centimeters. The correct choice is H.
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