-
Chemistry The time required for a certain chemical reaction is related to the amount of catalyst present during the reaction. The domain of the relation is the number of grains of catalyst, and the range is the number of seconds required for a fixed amount of the chemical to react. The table shows the data from several reactions.
Catalyst and Reaction Time
Number of Grains |
Number of Seconds |
2.0 |
180 |
2.5 |
6 |
2.7 |
0.05 |
2.9 |
0.001 |
3.0 |
6 |
3.1 |
15 |
3.2 |
37 |
3.3 |
176 |
- Is the relation a function?
- If the domain and range were interchanged, would the relation be a function? Explain.
Standardized Test Prep
SAT/ACT
- If
f
(
x
)
=
−
3
x
+
7
f open x close equals negative 3 x plus 7 and
g
(
x
)
=
−
7
x
+
3
,
g open x close equals negative 7 x plus 3 comma what is the value of
f
(
−
3
)
−
g
(
3
)
?
f open negative 3 close minus g open 3 close question mark
- 40
- 34
- 8
-
−
8
negative 8
- What is the formula for the volume of a cylinder,
V
=
π
r
2
h
,
v equals pi , r squared , h comma solved for h?
-
h
=
r
2
π
V
h equals . fraction r squared , over pi v end fraction
-
h
=
π
V
r
2
h equals . fraction pi v , over r squared end fraction
-
h
=
V
π
r
2
h equals . fraction v , over pi , r squared end fraction
-
h
=
π
r
2
V
h equals . fraction pi , r squared , over v end fraction
- Which of the following statements are true?
-
−
(
−
6
)
=
6
and
−
(
−
4
)
>
−
4
negative open negative 6 close equals 6 , and , minus open negative 4 close greater than negative 4
-
−
(
−
4
)
<
4
or
−
10
>
10
−
10
negative open negative 4 close less than 4 , or , minus 10 greater than 10 minus 10
-
5
+
6
=
11
or
9
−
2
=
11
5 plus 6 equals 11 , or , 9 minus 2 equals 11
-
17
>
2
or
6
<
9
17 greater than 2 , or , 6 less than 9
- I and II only
- I, II, and III only
- I, III, and IV only
- III and IV only
Short Response
- What are the numbers 1.9,
5
4
,
−
1.2
,
5 fourths , comma negative 1.2 comma and
3
square root of 3 in order from greatest to least?
Mixed Review
See Lessons 1-5 and 1-6.
Solve each equation or inequality.
- |3x + 9| = 11
-
19
+
|
x
−
1
|
=
33
19 plus vertical line x minus 1 vertical line equals 33
-
2
−
3
x
<
11
2 minus 3 x less than 11
-
5
x
−
3
≤
12
−
5
x
5 x minus 3 less than or equal to 12 minus 5 x
-
|
2
x
|
+
4
<
7
vertical line 2 x vertical line plus 4 less than 7
-
4
x
+
6
≥
−
6
4 x plus 6 greater than or equal to negative 6
See Lesson 1-4.
Get Ready! To prepare for Lesson 2-2, do Exercises 50–52.
Solve each equation for y.
- 12y = 3x
-
−
10
y
=
5
x
negative 10 y equals 5 x
-
3
4
y
=
15
x
3 fourths , y equals 15 x