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Math Problems
Algebra 2
Evaluate functions
{
f
(
1
)
=
−
16
f
(
n
)
=
f
(
n
−
1
)
⋅
(
−
1
2
)
f
(
3
)
=
□
\begin{array}{l}\left\{\begin{array}{l}f(1)=-16 \\ f(n)=f(n-1) \cdot\left(-\frac{1}{2}\right)\end{array}\right. \\ f(3)=\square\end{array}
{
f
(
1
)
=
−
16
f
(
n
)
=
f
(
n
−
1
)
⋅
(
−
2
1
)
f
(
3
)
=
□
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What is the sign of
a
⋅
(
−
b
b
)
a \cdot\left(\frac{-b}{b}\right)
a
⋅
(
b
−
b
)
when
a
=
0
a=0
a
=
0
and
b
<
0
b<0
b
<
0
?
\newline
Choose
1
1
1
answer:
\newline
(A) Positive
\newline
(B) Negative
\newline
(c) Zero
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What is the sign of
−
x
−
y
\frac{-x}{-y}
−
y
−
x
when
x
>
0
x>0
x
>
0
and
y
>
0
y>0
y
>
0
?
\newline
Choose
1
1
1
answer:
\newline
(A) Positive
\newline
(B) Negative
\newline
(c) Zero
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What is the sign of
−
4
⋅
(
3
−
4
)
-4 \cdot\left(\frac{3}{-4}\right)
−
4
⋅
(
−
4
3
)
?
\newline
Choose
1
1
1
answer:
\newline
(A) Positive
\newline
B Negative
\newline
(C) Zero
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What is the sign of
9
⋅
(
−
0.5
)
?
9 \cdot(-0.5) ?
9
⋅
(
−
0.5
)?
\newline
Choose
1
1
1
answer:
\newline
(A) Positive
\newline
(B) Negative
\newline
(C) Zero
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What is the sign of
−
9
⋅
(
0
−
3
)
-9 \cdot\left(\frac{0}{-3}\right)
−
9
⋅
(
−
3
0
)
?
\newline
Choose
1
1
1
answer:
\newline
(A) Positive
\newline
(B) Negative
\newline
(c) Zero
Get tutor help
What is the sign of
−
5
⋅
(
−
1
−
2
)
-5 \cdot\left(\frac{-1}{-2}\right)
−
5
⋅
(
−
2
−
1
)
?
\newline
Choose
1
1
1
answer:
\newline
(A) Positive
\newline
(B) Negative
\newline
(c) Zero
Get tutor help
What is the sign of
p
⋅
q
p
p \cdot \frac{q}{p}
p
⋅
p
q
when
p
>
0
p>0
p
>
0
and
q
=
0
q=0
q
=
0
?
\newline
Choose
1
1
1
answer:
\newline
(A) Positive
\newline
(B) Negative
\newline
(c) Zero
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d
y
d
x
=
6
x
\frac{d y}{d x}=6 x
d
x
d
y
=
6
x
and
y
(
2
)
=
8
y(2)=8
y
(
2
)
=
8
.
\newline
y
(
4
)
=
y(4)=
y
(
4
)
=
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f
′
(
x
)
=
8
x
3
−
12
x
f^{\prime}(x)=8 x^{3}-12 x
f
′
(
x
)
=
8
x
3
−
12
x
and
f
(
1
)
=
1
f(1)=1
f
(
1
)
=
1
.
\newline
f
(
−
2
)
=
f(-2)=
f
(
−
2
)
=
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Let
h
(
x
)
=
ln
(
x
)
cos
(
x
)
h(x)=\ln (x) \cos (x)
h
(
x
)
=
ln
(
x
)
cos
(
x
)
.
\newline
h
′
(
x
)
=
h^{\prime}(x)=
h
′
(
x
)
=
Get tutor help
Use the following function rule to find
f
(
6
)
f(6)
f
(
6
)
.
\newline
f
(
x
)
=
−
5
−
24
x
f(x) = -5 - \frac{24}{x}
f
(
x
)
=
−
5
−
x
24
\newline
f
(
6
)
=
f(6) =
f
(
6
)
=
_____
Get tutor help
Use the following function rule to find
f
(
140
)
f(140)
f
(
140
)
.
\newline
f
(
x
)
=
8
+
x
5
f(x) = 8 + \frac{x}{5}
f
(
x
)
=
8
+
5
x
\newline
f
(
140
)
=
f(140) =
f
(
140
)
=
_____
Get tutor help
Use the following function rule to find
f
(
35
)
f(35)
f
(
35
)
.
\newline
f
(
x
)
=
12
+
140
x
f(x) = 12 + \frac{140}{x}
f
(
x
)
=
12
+
x
140
\newline
f
(
35
)
=
f(35) =
f
(
35
)
=
_____
Get tutor help
Use the following function rule to find
f
(
6
)
f(6)
f
(
6
)
.
\newline
f
(
x
)
=
−
12
x
+
2
f(x) = \frac{-12}{x} + 2
f
(
x
)
=
x
−
12
+
2
\newline
f
(
6
)
=
f(6) =
f
(
6
)
=
_____
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Use the following function rule to find
f
(
137
)
f(137)
f
(
137
)
.
\newline
f
(
x
)
=
5
+
137
x
f(x) = 5 + \frac{137}{x}
f
(
x
)
=
5
+
x
137
\newline
f
(
137
)
=
f(137) =
f
(
137
)
=
_____
Get tutor help
Use the following function rule to find
f
(
12
)
f(12)
f
(
12
)
.
\newline
f
(
x
)
=
12
x
+
8
f(x) = \frac{12}{x} + 8
f
(
x
)
=
x
12
+
8
\newline
f
(
12
)
=
f(12) =
f
(
12
)
=
_____
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Use the following function rule to find
f
(
2
)
f(2)
f
(
2
)
.
\newline
f
(
x
)
=
3
−
72
x
f(x) = 3 - \frac{72}{x}
f
(
x
)
=
3
−
x
72
\newline
f
(
2
)
=
f(2) =
f
(
2
)
=
_____
Get tutor help
Use the following function rule to find
f
(
15
)
f(15)
f
(
15
)
.
\newline
f
(
x
)
=
−
120
x
+
6
f(x) = \frac{-120}{x} + 6
f
(
x
)
=
x
−
120
+
6
\newline
f
(
15
)
=
f(15) =
f
(
15
)
=
_____
Get tutor help
Use the following function rule to find
f
(
−
2
)
f(-2)
f
(
−
2
)
.
\newline
f
(
x
)
=
−
10
x
+
1
f(x) = -10x + 1
f
(
x
)
=
−
10
x
+
1
\newline
f
(
−
2
)
=
f(-2) =
f
(
−
2
)
=
_____
Get tutor help
Use the following function rule to find
f
(
12
)
f(12)
f
(
12
)
.
\newline
f
(
x
)
=
6
−
96
x
f(x) = 6 - \frac{96}{x}
f
(
x
)
=
6
−
x
96
\newline
f
(
12
)
=
f(12) =
f
(
12
)
=
_____
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Use the following function rule to find
f
(
86
)
f(86)
f
(
86
)
.
\newline
f
(
x
)
=
2
−
86
x
f(x) = 2 - \frac{86}{x}
f
(
x
)
=
2
−
x
86
\newline
f
(
86
)
=
f(86) =
f
(
86
)
=
_____
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Use the following function rule to find
f
(
6
)
f(6)
f
(
6
)
.
\newline
f
(
x
)
=
−
8
−
36
x
f(x) = -8 - \frac{36}{x}
f
(
x
)
=
−
8
−
x
36
\newline
f
(
6
)
=
f(6) =
f
(
6
)
=
_____
Get tutor help
f)
{
x
+
y
2
−
x
−
y
3
=
3
x
+
2
y
3
−
x
−
2
y
4
=
3
\left\{\begin{array}{l}\frac{x+y}{2}-\frac{x-y}{3}=3 \\ \frac{x+2 y}{3}-\frac{x-2 y}{4}=3\end{array}\right.
{
2
x
+
y
−
3
x
−
y
=
3
3
x
+
2
y
−
4
x
−
2
y
=
3
Get tutor help
{
f
(
1
)
=
0
f
(
n
)
=
f
(
n
−
1
)
+
2
\left\{\begin{array}{l} f(1)=0 \\ f(n)=f(n-1)+2 \end{array}\right.
{
f
(
1
)
=
0
f
(
n
)
=
f
(
n
−
1
)
+
2
\newline
Find an explicit formula for
f
(
n
)
f(n)
f
(
n
)
.
\newline
f
(
n
)
=
f(n)=
f
(
n
)
=
\newline
______________
Get tutor help
If
f
(
1
)
=
1
f(1)=1
f
(
1
)
=
1
and
f
(
n
)
=
f
(
n
−
1
)
−
1
f(n)=f(n-1)-1
f
(
n
)
=
f
(
n
−
1
)
−
1
then find the value of
f
(
5
)
f(5)
f
(
5
)
.
\newline
Answer:
Get tutor help
If
f
(
1
)
=
4
f(1)=4
f
(
1
)
=
4
and
f
(
n
)
=
−
2
f
(
n
−
1
)
f(n)=-2 f(n-1)
f
(
n
)
=
−
2
f
(
n
−
1
)
then find the value of
f
(
4
)
f(4)
f
(
4
)
.
\newline
Answer:
Get tutor help
What is the slope of the line?
\newline
5
(
y
+
2
)
=
4
(
x
−
3
)
5(y+2)=4(x-3)
5
(
y
+
2
)
=
4
(
x
−
3
)
\newline
Choose
1
1
1
answer:
\newline
(A)
2
3
\frac{2}{3}
3
2
\newline
(B)
4
5
\frac{4}{5}
5
4
\newline
(C)
5
4
\frac{5}{4}
4
5
\newline
(D)
3
2
\frac{3}{2}
2
3
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f
(
x
)
=
x
4
+
4
x
3
−
7
x
2
−
22
x
+
24
f(x)=x^{4}+4 x^{3}-7 x^{2}-22 x+24
f
(
x
)
=
x
4
+
4
x
3
−
7
x
2
−
22
x
+
24
\newline
The function
f
f
f
is shown. If
x
+
3
x+3
x
+
3
is a factor of
f
f
f
, what is the value of
f
(
−
3
)
f(-3)
f
(
−
3
)
?
\newline
Choose
1
1
1
answer:
\newline
(A)
−
3
-3
−
3
\newline
(B)
0
0
0
\newline
(C)
3
3
3
\newline
(D)
24
24
24
Get tutor help
Convert to a fraction in simplest form:
\newline
2.388888
…
2.388888 \ldots
2.388888
…
\newline
2
7
18
2 \frac{7}{18}
2
18
7
\newline
2
38
99
2 \frac{38}{99}
2
99
38
\newline
2
3
8
2 \frac{3}{8}
2
8
3
\newline
2
38
100
2 \frac{38}{100}
2
100
38
\newline
2
38
90
2 \frac{38}{90}
2
90
38
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If
x
y
=
3
8
\frac{x}{y}=\frac{3}{8}
y
x
=
8
3
and
y
z
=
2
15
\frac{y}{z}=\frac{2}{15}
z
y
=
15
2
\newline
Then
x
z
=
\frac{x}{z}=
z
x
=
Get tutor help
Use the following function rule to find
h
(
3
j
)
h(3j)
h
(
3
j
)
. Simplify your answer.
\newline
h
(
m
)
=
−
7
m
h(m) = -7m
h
(
m
)
=
−
7
m
\newline
h
(
3
j
)
=
h(3j) =
h
(
3
j
)
=
______
Get tutor help
Use the following function rule to find
f
(
a
−
7
)
f(a - 7)
f
(
a
−
7
)
. Simplify your answer.
\newline
f
(
x
)
=
−
2
x
f(x) = -2x
f
(
x
)
=
−
2
x
\newline
f
(
a
−
7
)
=
f(a - 7) =
f
(
a
−
7
)
=
______
Get tutor help
Use the following function rule to find
f
(
−
9
q
)
f(-9q)
f
(
−
9
q
)
. Simplify your answer.
\newline
f
(
j
)
=
8
j
f(j) = 8j
f
(
j
)
=
8
j
\newline
f
(
−
9
q
)
=
f(-9q) =
f
(
−
9
q
)
=
_
_
\_\_
__
Get tutor help
If
x
=
2
sin
θ
−
sin
2
θ
x=2\sin \theta-\sin 2\theta
x
=
2
sin
θ
−
sin
2
θ
and
y
=
2
cos
θ
−
cos
2
θ
y=2\cos \theta-\cos 2\theta
y
=
2
cos
θ
−
cos
2
θ
,
θ
∈
[
0
,
2
π
]
\theta \in[0,2\pi]
θ
∈
[
0
,
2
π
]
, then
(
d
2
y
)
/
(
d
x
2
)
(d^{2}y)/(dx^{2})
(
d
2
y
)
/
(
d
x
2
)
at
θ
=
π
\theta=\pi
θ
=
π
is :
Get tutor help
Aileen had
3
5
\frac{3}{5}
5
3
as many beads as Betty. When Aileen gave Betty some beads, the number of beads Aileen had to the number of beads Betty had become
1
:
3
1: 3
1
:
3
. Betty then gave
2
3
\frac{2}{3}
3
2
of her beads away and had
30
30
30
beads left. How many beads did Aileen have at first?
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Prove that
Tan
A
sec
A
−
1
−
1
+
Tan
A
sec
A
+
1
=
2
cosec
θ
\frac{\operatorname{Tan} A}{\sec A-1}-\frac{1+\operatorname{Tan} A}{\sec A+1}=2 \operatorname{cosec} \theta
s
e
c
A
−
1
Tan
A
−
s
e
c
A
+
1
1
+
Tan
A
=
2
cosec
θ
Get tutor help
Let
A
=
{
1
,
2
,
3
}
A = \{1, 2, 3\}
A
=
{
1
,
2
,
3
}
and
B
=
{
2
,
4
}
B = \{2, 4\}
B
=
{
2
,
4
}
. What is
A
∪
B
A \cup B
A
∪
B
?
\newline
Choices:
\newline
(A)
{
1
,
3
,
4
}
\{1, 3, 4\}
{
1
,
3
,
4
}
\newline
(B)
{
1
,
2
,
3
,
4
}
\{1, 2, 3, 4\}
{
1
,
2
,
3
,
4
}
\newline
(C)
{
4
}
\{4\}
{
4
}
\newline
(D)
{
2
}
\{2\}
{
2
}
Get tutor help
Use the following function rule to find
f
(
144
)
f(144)
f
(
144
)
.
\newline
f
(
x
)
=
5
x
+
7
f(x) = 5\sqrt{x} + 7
f
(
x
)
=
5
x
+
7
\newline
f
(
144
)
=
f(144) =
f
(
144
)
=
_____
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Suppose that
sin
α
=
3
13
\sin \alpha=\frac{3}{\sqrt{13}}
sin
α
=
13
3
and
0
∘
<
α
<
9
0
∘
0^{\circ}<\alpha<90^{\circ}
0
∘
<
α
<
9
0
∘
\newline
Find the exact values of
cos
α
2
\cos \frac{\alpha}{2}
cos
2
α
and
tan
α
2
\tan \frac{\alpha}{2}
tan
2
α
.
\newline
cos
α
2
=
\cos \frac{\alpha}{2}=
cos
2
α
=
\newline
tan
α
2
=
\tan \frac{\alpha}{2}=
tan
2
α
=
Get tutor help
Jose bought
f
f
f
folders that cost
$
0.35
\$ 0.35
$0.35
each and
m
m
m
markers that cost
$
0.25
\$ 0.25
$0.25
each. Sales tax is
9
%
9 \%
9%
. Which expression represents the total amount Jose paid, including tax?
\newline
1.09
(
f
+
m
)
1.09(f+m)
1.09
(
f
+
m
)
\newline
0.35
(
0.09
f
)
+
0.25
(
0.09
m
)
0.35(0.09 f)+0.25(0.09 m)
0.35
(
0.09
f
)
+
0.25
(
0.09
m
)
\newline
(
0.35
f
+
0.25
m
)
+
0.09
(
0.35
f
+
0.25
m
)
(0.35 f+0.25 m)+0.09(0.35 f+0.25 m)
(
0.35
f
+
0.25
m
)
+
0.09
(
0.35
f
+
0.25
m
)
\newline
1.09
(
0.35
+
0.25
)
1.09(0.35+0.25)
1.09
(
0.35
+
0.25
)
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Use the following function rule to find
f
(
6
)
f(6)
f
(
6
)
.
\newline
f
(
x
)
=
6
x
+
1
f(x) = 6x + 1
f
(
x
)
=
6
x
+
1
\newline
f
(
6
)
=
‾
f(6) = \underline{\hspace{2em}}
f
(
6
)
=
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Use the following function rule to find
f
(
3
)
f(3)
f
(
3
)
.
\newline
f
(
x
)
=
7
−
75
x
f
(
3
)
=
□
\begin{array}{l} f(x)=7-\frac{75}{x} \\ f(3)=\square \end{array}
f
(
x
)
=
7
−
x
75
f
(
3
)
=
□
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If
f
(
x
)
=
(
2
x
−
5
)
(
x
+
2
)
(
x
+
6
)
+
8
f(x)=(2 x-5)(x+2)(x+6)+8
f
(
x
)
=
(
2
x
−
5
)
(
x
+
2
)
(
x
+
6
)
+
8
, what is the value of
f
(
−
6
)
?
f(-6) ?
f
(
−
6
)?
\newline
Choose
1
1
1
answer:
\newline
(A)
−
60
-60
−
60
\newline
(B)
0
0
0
\newline
(C)
2
2
2
\newline
(D)
8
8
8
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f
(
x
)
=
2
x
5
+
x
4
−
18
x
3
−
17
x
2
+
20
x
+
12
f(x)=2 x^{5}+x^{4}-18 x^{3}-17 x^{2}+20 x+12
f
(
x
)
=
2
x
5
+
x
4
−
18
x
3
−
17
x
2
+
20
x
+
12
\newline
The function
f
f
f
is shown. If
x
−
3
x-3
x
−
3
is a factor of
f
f
f
, what is the value of
f
(
3
)
f(3)
f
(
3
)
?
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Let
f
(
x
)
=
4
x
−
15
f(x)=4 x-15
f
(
x
)
=
4
x
−
15
. Find each of the following:
\newline
f
(
a
)
=
2
f
(
a
)
=
f
(
2
a
)
=
\begin{array}{l} f(a)= \\ 2 f(a)= \\ f(2 a)= \end{array}
f
(
a
)
=
2
f
(
a
)
=
f
(
2
a
)
=
\newline
f
(
a
+
2
)
=
f(a+2)=
f
(
a
+
2
)
=
\newline
f
(
a
)
+
f
(
2
)
=
f(a)+f(2)=
f
(
a
)
+
f
(
2
)
=
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f
(
x
)
=
x
2
+
1
f(x) = x^2 + 1
f
(
x
)
=
x
2
+
1
\newline
g
(
x
)
=
5
–
x
g(x) = 5 – x
g
(
x
)
=
5–
x
\newline
Find
(
f
+
g
)
(
x
)
=
(f + g)(x) =
(
f
+
g
)
(
x
)
=
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Find an explicit formula for the geometric sequence.
\newline
120
,
60
,
30
,
15
,
…
120, 60, 30, 15, \dots
120
,
60
,
30
,
15
,
…
\newline
Note: the first term should be
a
(
1
)
a(1)
a
(
1
)
.
\newline
a
(
n
)
=
□
a(n)=\square
a
(
n
)
=
□
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Daniela has a three-season acting contract for a new TV show. Her pay for Season
1
1
1
is
p
p
p
dollars. Then, her pay will increase by
5
%
5\%
5%
from Season
1
1
1
to Season
2
2
2
and by
10
%
10\%
10%
from Season
2
2
2
to Season
3
3
3
. Which of the following represents Daniela's Season
3
3
3
pay in terms of
p
p
p
?
\newline
Choose
1
1
1
answer:
\newline
(A)
1.15
p
1.15p
1.15
p
\newline
(B)
p
1.15
\frac{p}{1.15}
1.15
p
\newline
(C)
(
1.05
)
(
1.1
)
p
(1.05)(1.1)p
(
1.05
)
(
1.1
)
p
\newline
(D)
p
(
1.05
)
(
1.1
)
\frac{p}{(1.05)(1.1)}
(
1.05
)
(
1.1
)
p
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If
f
(
1
)
=
7
f(1)=7
f
(
1
)
=
7
and
f
(
n
)
=
4
f
(
n
−
1
)
+
n
f(n)=4 f(n-1)+n
f
(
n
)
=
4
f
(
n
−
1
)
+
n
then find the value of
f
(
3
)
f(3)
f
(
3
)
.
\newline
Answer:
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1
2
3
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