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Math Problems
Algebra 1
Compare linear and exponential growth
∑
k
=
0
99
2
(
3
)
k
≈
\sum_{k=0}^{99} 2(3)^{k} \approx
k
=
0
∑
99
2
(
3
)
k
≈
\newline
Choose
1
1
1
answer:
\newline
(A)
5.15
⋅
1
0
47
5.15 \cdot 10^{47}
5.15
⋅
1
0
47
\newline
(B)
3.80
⋅
1
0
30
3.80 \cdot 10^{30}
3.80
⋅
1
0
30
\newline
(C)
2.37
⋅
1
0
−
30
2.37 \cdot 10^{-30}
2.37
⋅
1
0
−
30
\newline
(D)
7.76
⋅
1
0
−
48
7.76 \cdot 10^{-48}
7.76
⋅
1
0
−
48
Get tutor help
Consider the complex number
z
=
5
(
cos
(
1
5
∘
)
+
i
sin
(
1
5
∘
)
)
z=5\left(\cos \left(15^{\circ}\right)+i \sin \left(15^{\circ}\right)\right)
z
=
5
(
cos
(
1
5
∘
)
+
i
sin
(
1
5
∘
)
)
.
\newline
Which of the following complex numbers best approximates
z
z
z
?
\newline
Choose
1
1
1
answer:
\newline
(A)
4.8
+
1.3
i
4.8+1.3 i
4.8
+
1.3
i
\newline
(B)
1.3
+
4.8
i
1.3+4.8 i
1.3
+
4.8
i
\newline
(C)
−
1.3
+
4.8
i
-1.3+4.8 i
−
1.3
+
4.8
i
\newline
(D)
−
4.8
+
1.3
i
-4.8+1.3 i
−
4.8
+
1.3
i
Get tutor help
A complex number
z
1
z_{1}
z
1
has a magnitude
∣
z
1
∣
=
3
\left|z_{1}\right|=3
∣
z
1
∣
=
3
and an angle
θ
1
=
2
0
∘
\theta_{1}=20^{\circ}
θ
1
=
2
0
∘
.
\newline
Express
z
1
z_{1}
z
1
in rectangular form, as
z
1
=
a
+
b
i
z_{1}=a+b i
z
1
=
a
+
bi
.
\newline
Round
a
a
a
and
b
b
b
to the nearest thousandth.
\newline
z
1
=
□
+
□
i
z_{1}=\square+\square i
z
1
=
□
+
□
i
Get tutor help
A complex number
z
1
z_{1}
z
1
has a magnitude
∣
z
1
∣
=
9
\left|z_{1}\right|=9
∣
z
1
∣
=
9
and an angle
θ
1
=
5
0
∘
\theta_{1}=50^{\circ}
θ
1
=
5
0
∘
.
\newline
Express
z
1
z_{1}
z
1
in rectangular form, as
z
1
=
a
+
b
i
z_{1}=a+b i
z
1
=
a
+
bi
.
\newline
Round
a
a
a
and
b
b
b
to the nearest thousandth.
\newline
z
1
=
□
+
□
i
z_{1}=\square+\square i
z
1
=
□
+
□
i
Get tutor help
Consider the complex number
\newline
z
=
−
5
2
2
+
5
2
2
i
z=-\frac{5 \sqrt{2}}{2}+\frac{5 \sqrt{2}}{2} i \text { }
z
=
−
2
5
2
+
2
5
2
i
\newline
Which of the following complex numbers best approximates
z
3
z^{3}
z
3
? Hint:
z
z
z
has a modulus of
5
5
5
and an argument of
13
5
∘
135^{\circ}
13
5
∘
.
\newline
Choose
1
1
1
answer:
\newline
(A)
−
10.6
+
10.6
i
-10.6+10.6 i
−
10.6
+
10.6
i
\newline
(B)
−
125
-125
−
125
\newline
(C)
−
125
i
-125 i
−
125
i
\newline
(D)
88.4
+
88.4
i
88.4+88.4 i
88.4
+
88.4
i
Get tutor help
A complex number
z
1
z_{1}
z
1
has a magnitude
∣
z
1
∣
=
7
\left|z_{1}\right|=7
∣
z
1
∣
=
7
and an angle
θ
1
=
15
8
∘
\theta_{1}=158^{\circ}
θ
1
=
15
8
∘
.
\newline
Express
z
1
z_{1}
z
1
in rectangular form, as
z
1
=
a
+
b
i
z_{1}=a+b i
z
1
=
a
+
bi
.
\newline
Round
a
a
a
and
b
b
b
to the nearest thousandth.
\newline
z
1
=
□
+
□
i
z_{1}=\square+\square i
z
1
=
□
+
□
i
Get tutor help
A complex number
z
1
z_{1}
z
1
has a magnitude
∣
z
1
∣
=
22
\left|z_{1}\right|=22
∣
z
1
∣
=
22
and an angle
θ
1
=
5
6
∘
\theta_{1}=56^{\circ}
θ
1
=
5
6
∘
.
\newline
Express
z
1
z_{1}
z
1
in rectangular form, as
z
1
=
a
+
b
i
z_{1}=a+b i
z
1
=
a
+
bi
.
\newline
Round
a
a
a
and
b
b
b
to the nearest thousandth.
\newline
z
1
=
□
+
□
i
z_{1}=\square+\square i
z
1
=
□
+
□
i
Get tutor help
A complex number
z
1
z_{1}
z
1
has a magnitude
∣
z
1
∣
=
18
\left|z_{1}\right|=18
∣
z
1
∣
=
18
and an angle
θ
1
=
12
2
∘
\theta_{1}=122^{\circ}
θ
1
=
12
2
∘
.
\newline
Express
z
1
z_{1}
z
1
in rectangular form, as
z
1
=
a
+
b
i
z_{1}=a+b i
z
1
=
a
+
bi
.
\newline
Round
a
a
a
and
b
b
b
to the nearest thousandth.
\newline
z
1
=
□
+
□
i
z_{1}=\square+\square i
z
1
=
□
+
□
i
Get tutor help
A complex number
z
1
z_{1}
z
1
has a magnitude
∣
z
1
∣
=
12
\left|z_{1}\right|=12
∣
z
1
∣
=
12
and an angle
θ
1
=
16
3
∘
\theta_{1}=163^{\circ}
θ
1
=
16
3
∘
.
\newline
Express
z
1
z_{1}
z
1
in rectangular form, as
z
1
=
a
+
b
i
z_{1}=a+b i
z
1
=
a
+
bi
.
\newline
Round
a
a
a
and
b
b
b
to the nearest thousandth.
\newline
z
1
=
□
+
□
i
z_{1}=\square+\square i
z
1
=
□
+
□
i
Get tutor help
Find the limit as
x
x
x
approaches negative infinity.
\newline
lim
x
→
−
∞
16
x
6
−
x
2
6
x
3
+
x
2
=
\lim _{x \rightarrow-\infty} \frac{\sqrt{16 x^{6}-x^{2}}}{6 x^{3}+x^{2}}=
x
→
−
∞
lim
6
x
3
+
x
2
16
x
6
−
x
2
=
Get tutor help
These are the component forms of vectors
e
⃗
\vec{e}
e
and
f
⃗
\vec{f}
f
:
\newline
e
⃗
=
(
3
,
5
)
f
⃗
=
(
1
,
−
6
)
\begin{array}{l} \vec{e}=(3,5) \\ \vec{f}=(1,-6) \end{array}
e
=
(
3
,
5
)
f
=
(
1
,
−
6
)
\newline
Add the vectors.
\newline
e
⃗
+
f
⃗
=
(
□
,
□
)
\vec{e}+\vec{f}=(\square, \square)
e
+
f
=
(
□
,
□
)
Get tutor help
These are the component forms of vectors
u
⃗
\vec{u}
u
and
w
⃗
\vec{w}
w
:
\newline
u
⃗
=
(
−
4
,
−
3
)
w
⃗
=
(
−
1
,
5
)
\begin{aligned} \vec{u} & =(-4,-3) \\ \vec{w} & =(-1,5) \end{aligned}
u
w
=
(
−
4
,
−
3
)
=
(
−
1
,
5
)
\newline
Add the vectors.
\newline
u
⃗
+
w
⃗
=
(
□
,
□
)
\vec{u}+\vec{w}=(\square, \square)
u
+
w
=
(
□
,
□
)
Get tutor help
These are the component forms of vectors
p
⃗
\vec{p}
p
and
q
⃗
\vec{q}
q
:
\newline
p
⃗
=
(
2
,
−
2
)
q
⃗
=
(
−
3
,
−
1
)
\begin{array}{l} \vec{p}=(2,-2) \\ \vec{q}=(-3,-1) \end{array}
p
=
(
2
,
−
2
)
q
=
(
−
3
,
−
1
)
\newline
Add the vectors.
\newline
p
⃗
+
q
⃗
=
(
□
,
□
)
\vec{p}+\vec{q}=(\square, \square)
p
+
q
=
(
□
,
□
)
Get tutor help
f
(
x
)
=
x
3
−
6
f(x)=x^{3}-6
f
(
x
)
=
x
3
−
6
\newline
h
(
x
)
=
2
x
−
15
3
h(x)=\sqrt[3]{2 x-15}
h
(
x
)
=
3
2
x
−
15
\newline
Write
f
(
h
(
x
)
)
f(h(x))
f
(
h
(
x
))
as an expression in terms of
x
x
x
.
\newline
f
(
h
(
x
)
)
=
f(h(x))=
f
(
h
(
x
))
=
Get tutor help
f
(
x
)
=
x
−
5
f(x)=x-5
f
(
x
)
=
x
−
5
\newline
g
(
x
)
=
x
2
−
12
x
2
+
4
g(x)=\frac{x^{2}-12}{x^{2}+4}
g
(
x
)
=
x
2
+
4
x
2
−
12
\newline
Write
g
(
f
(
x
)
)
g(f(x))
g
(
f
(
x
))
as an expression in terms of
x
x
x
.
\newline
g
(
f
(
x
)
)
=
g(f(x))=
g
(
f
(
x
))
=
Get tutor help
f
(
x
)
=
3
⋅
2
x
h
(
x
)
=
2
x
−
7
\begin{array}{l} f(x)=3 \cdot 2^{x} \\ h(x)=2 x-7 \end{array}
f
(
x
)
=
3
⋅
2
x
h
(
x
)
=
2
x
−
7
\newline
Evaluate.
\newline
h
(
f
(
2
)
)
=
h(f(2))=
h
(
f
(
2
))
=
Get tutor help
g
(
x
)
=
9
+
5
4
x
h
(
x
)
=
10
x
−
28
\begin{array}{l} g(x)=9+\frac{5}{4} x \\ h(x)=10 x-28 \end{array}
g
(
x
)
=
9
+
4
5
x
h
(
x
)
=
10
x
−
28
\newline
Write
(
g
∘
h
)
(
x
)
(g \circ h)(x)
(
g
∘
h
)
(
x
)
as an expression in terms of
x
x
x
.
\newline
(
g
∘
h
)
(
x
)
=
(g \circ h)(x)=
(
g
∘
h
)
(
x
)
=
Get tutor help
g
(
y
)
=
y
−
15
y
h
(
y
)
=
−
3
y
\begin{array}{l} g(y)=\frac{y-15}{y} \\ h(y)=-3 y \end{array}
g
(
y
)
=
y
y
−
15
h
(
y
)
=
−
3
y
\newline
Evaluate.
\newline
h
(
g
(
5
)
)
=
h(g(5))=
h
(
g
(
5
))
=
Get tutor help
g
(
x
)
=
−
20
−
3
x
g(x)=-20-3 x
g
(
x
)
=
−
20
−
3
x
\newline
h
(
x
)
=
(
1
2
)
x
h(x)=\left(\frac{1}{2}\right)^{x}
h
(
x
)
=
(
2
1
)
x
\newline
Evaluate.
\newline
(
g
∘
h
)
(
−
2
)
=
(g \circ h)(-2)=
(
g
∘
h
)
(
−
2
)
=
Get tutor help
g
(
b
)
=
5
b
−
9
g(b)=5 b-9
g
(
b
)
=
5
b
−
9
\newline
h
(
b
)
=
(
b
−
1
)
2
h(b)=(b-1)^{2}
h
(
b
)
=
(
b
−
1
)
2
\newline
Evaluate.
\newline
(
h
∘
g
)
(
−
6
)
=
(h \circ g)(-6)=
(
h
∘
g
)
(
−
6
)
=
Get tutor help
f
(
x
)
=
2
x
+
3
f(x)=2 x+3
f
(
x
)
=
2
x
+
3
\newline
g
(
x
)
=
x
2
−
3
x
+
1
g(x)=x^{2}-3 x+1
g
(
x
)
=
x
2
−
3
x
+
1
\newline
Write
g
(
f
(
x
)
)
g(f(x))
g
(
f
(
x
))
as an expression in terms of
x
x
x
.
\newline
g
(
f
(
x
)
)
=
g(f(x))=
g
(
f
(
x
))
=
Get tutor help
g
(
x
)
=
9
+
4
x
g(x)=9+4 x
g
(
x
)
=
9
+
4
x
\newline
h
(
x
)
=
x
+
21
5
h(x)=\frac{x+21}{5}
h
(
x
)
=
5
x
+
21
\newline
Write
(
h
∘
g
)
(
x
)
(h \circ g)(x)
(
h
∘
g
)
(
x
)
as an expression in terms of
x
x
x
.
\newline
(
h
∘
g
)
(
x
)
=
(h \circ g)(x)=
(
h
∘
g
)
(
x
)
=
Get tutor help
g
(
x
)
=
3
−
x
2
h
(
x
)
=
4
−
3
x
\begin{array}{l} g(x)=3-x^{2} \\ h(x)=4-3 x \end{array}
g
(
x
)
=
3
−
x
2
h
(
x
)
=
4
−
3
x
\newline
Write
(
g
∘
h
)
(
x
)
(g \circ h)(x)
(
g
∘
h
)
(
x
)
as an expression in terms of
x
x
x
.
\newline
(
g
∘
h
)
(
x
)
=
(g \circ h)(x)=
(
g
∘
h
)
(
x
)
=
Get tutor help
f
(
x
)
=
13
−
x
h
(
x
)
=
x
2
−
6
x
−
18
\begin{array}{l} f(x)=13-x \\ h(x)=x^{2}-6 x-18 \end{array}
f
(
x
)
=
13
−
x
h
(
x
)
=
x
2
−
6
x
−
18
\newline
Write
(
f
∘
h
)
(
x
)
(f \circ h)(x)
(
f
∘
h
)
(
x
)
as an expression in terms of
x
x
x
.
\newline
(
f
∘
h
)
(
x
)
=
(f \circ h)(x)=
(
f
∘
h
)
(
x
)
=
Get tutor help
Skyler climbs a watchtower to guard against forest fires.
\newline
The function
R
(
h
)
=
13
h
R(h)=\sqrt{13 h}
R
(
h
)
=
13
h
gives the distance, in kilometers, that Skyler can see when their eyes are
h
h
h
meters above the ground.
\newline
The function
A
(
d
)
=
π
d
2
A(d)=\pi d^{2}
A
(
d
)
=
π
d
2
gives the area, in square kilometers, that Skyler can guard when they can see a distance of
d
d
d
kilometers.
\newline
Which expression models the area Skyler can guard when their eyes are
h
h
h
meters above the ground?
\newline
Choose
1
1
1
answer:
\newline
(A)
13
π
h
13 \pi h
13
πh
\newline
(B)
13
π
h
2
13 \pi h^{2}
13
π
h
2
\newline
(C)
13
π
h
2
\sqrt{13 \pi h^{2}}
13
π
h
2
\newline
(D)
169
π
h
2
169 \pi h^{2}
169
π
h
2
Get tutor help
f
(
a
)
=
(
a
+
2
)
2
g
(
a
)
=
5
a
+
4
\begin{array}{l} f(a)=(a+2)^{2} \\ g(a)=5 a+4 \end{array}
f
(
a
)
=
(
a
+
2
)
2
g
(
a
)
=
5
a
+
4
\newline
Evaluate.
\newline
g
(
f
(
−
2
)
)
=
g(f(-2))=
g
(
f
(
−
2
))
=
Get tutor help
g
(
m
)
=
m
3
g(m)=m^{3}
g
(
m
)
=
m
3
\newline
h
(
m
)
=
m
2
m
+
3
h(m)=\frac{m^{2}}{m+3}
h
(
m
)
=
m
+
3
m
2
\newline
Evaluate.
\newline
(
g
∘
h
)
(
6
)
=
(g \circ h)(6)=
(
g
∘
h
)
(
6
)
=
Get tutor help
f
(
z
)
=
(
1
3
)
z
h
(
z
)
=
z
+
4
z
−
2
\begin{array}{l} f(z)=\left(\frac{1}{3}\right)^{z} \\ h(z)=\frac{z+4}{z-2} \end{array}
f
(
z
)
=
(
3
1
)
z
h
(
z
)
=
z
−
2
z
+
4
\newline
Evaluate.
\newline
(
h
∘
f
)
(
0
)
=
(h \circ f)(0)=
(
h
∘
f
)
(
0
)
=
Get tutor help
f
(
t
)
=
t
t
−
8
h
(
t
)
=
−
2
t
\begin{array}{l} f(t)=\frac{t}{t-8} \\ h(t)=-2 t \end{array}
f
(
t
)
=
t
−
8
t
h
(
t
)
=
−
2
t
\newline
Evaluate.
\newline
f
(
h
(
−
5
)
)
=
f(h(-5))=
f
(
h
(
−
5
))
=
Get tutor help
f
(
x
)
=
(
3
x
−
5
)
3
h
(
x
)
=
2
x
3
+
8
\begin{array}{l} f(x)=(3 x-5)^{3} \\ h(x)=2 \sqrt[3]{x}+8 \end{array}
f
(
x
)
=
(
3
x
−
5
)
3
h
(
x
)
=
2
3
x
+
8
\newline
Write
h
(
f
(
x
)
)
h(f(x))
h
(
f
(
x
))
as an expression in terms of
x
x
x
.
\newline
h
(
f
(
x
)
)
=
h(f(x))=
h
(
f
(
x
))
=
Get tutor help
f
(
x
)
=
2
x
3
+
8
f(x)=2 x^{3}+8
f
(
x
)
=
2
x
3
+
8
\newline
h
(
x
)
=
12
−
5
x
3
h(x)=\sqrt[3]{12-5 x}
h
(
x
)
=
3
12
−
5
x
\newline
Write
(
f
∘
h
)
(
x
)
(f \circ h)(x)
(
f
∘
h
)
(
x
)
as an expression in terms of
x
x
x
.
\newline
(
f
∘
h
)
(
x
)
=
(f \circ h)(x)=
(
f
∘
h
)
(
x
)
=
Get tutor help
f
(
x
)
=
7
x
+
2
f(x)=7 x+2
f
(
x
)
=
7
x
+
2
\newline
g
(
x
)
=
x
2
x
−
5
g(x)=\frac{x^{2}}{x-5}
g
(
x
)
=
x
−
5
x
2
\newline
Write
(
g
∘
f
)
(
x
)
(g \circ f)(x)
(
g
∘
f
)
(
x
)
as an expression in terms of
x
x
x
.
\newline
(
g
∘
f
)
(
x
)
=
(g \circ f)(x)=
(
g
∘
f
)
(
x
)
=
Get tutor help
f
(
x
)
=
6
−
1
2
x
h
(
x
)
=
4
(
x
−
3
)
2
\begin{array}{l} f(x)=6-\frac{1}{2} x \\ h(x)=4(x-3)^{2} \end{array}
f
(
x
)
=
6
−
2
1
x
h
(
x
)
=
4
(
x
−
3
)
2
\newline
Write
f
(
h
(
x
)
)
f(h(x))
f
(
h
(
x
))
as an expression in terms of
x
x
x
.
\newline
f
(
h
(
x
)
)
=
f(h(x))=
f
(
h
(
x
))
=
Get tutor help
A passcode to enter a building is a sequence of
4
4
4
single digit numbers
(
0
−
9
)
(0-9)
(
0
−
9
)
, and repeated digits aren't allowed.
\newline
Suppose someone doesn't know the passcode and randomly guesses a sequence of
4
4
4
digits.
\newline
What is the probability that they guess the correct sequence?
\newline
Choose
1
1
1
answer:
\newline
(A)
1
10
P
4
\frac{1}{{ }_{10} \mathrm{P}_{4}}
10
P
4
1
\newline
(B)
1
10
C
4
\frac{1}{{ }_{10} \mathrm{C}_{4}}
10
C
4
1
\newline
(C)
4
P
4
10
P
4
\frac{{ }_{4} \mathrm{P}_{4}}{{ }_{10} \mathrm{P}_{4}}
10
P
4
4
P
4
\newline
(D)
(
4
P
2
)
⋅
(
4
P
2
)
10
P
4
\frac{\left({ }_{4} \mathrm{P}_{2}\right) \cdot\left({ }_{4} \mathrm{P}_{2}\right)}{{ }_{10} \mathrm{P}_{4}}
10
P
4
(
4
P
2
)
⋅
(
4
P
2
)
Get tutor help
h
(
t
)
=
50
−
t
5
h
(
35
)
=
\begin{array}{l}h(t)=50-\frac{t}{5} \\ h(35)=\end{array}
h
(
t
)
=
50
−
5
t
h
(
35
)
=
Get tutor help
4
⋅
2
−
t
5
=
288
4 \cdot 2^{-\frac{t}{5}}=288
4
⋅
2
−
5
t
=
288
\newline
Which of the following is the solution of the equation?
\newline
Choose
1
1
1
answer:
\newline
(A)
t
=
−
5
log
8
(
288
)
t=-5 \log _{8}(288)
t
=
−
5
lo
g
8
(
288
)
\newline
(B)
t
=
−
5
log
288
(
8
)
t=-5 \log _{288}(8)
t
=
−
5
lo
g
288
(
8
)
\newline
(C)
t
=
−
5
log
2
(
72
)
t=-5 \log _{2}(72)
t
=
−
5
lo
g
2
(
72
)
\newline
(D)
t
=
−
5
log
72
(
2
)
t=-5 \log _{72}(2)
t
=
−
5
lo
g
72
(
2
)
Get tutor help
3
⋅
2
x
5
=
150
3 \cdot 2^{\frac{x}{5}}=150
3
⋅
2
5
x
=
150
\newline
Which of the following is the solution of the equation?
\newline
Choose
1
1
1
answer:
\newline
(A)
x
=
log
50
(
2
)
x=\log _{50}(2)
x
=
lo
g
50
(
2
)
\newline
(B)
x
=
5
log
50
(
2
)
x=5 \log _{50}(2)
x
=
5
lo
g
50
(
2
)
\newline
(C)
x
=
5
log
2
(
50
)
x=5 \log _{2}(50)
x
=
5
lo
g
2
(
50
)
\newline
(D)
x
=
log
2
(
50
)
x=\log _{2}(50)
x
=
lo
g
2
(
50
)
Get tutor help
Consider the equation
\newline
−
16
⋅
1
0
6
x
=
−
80
.
-16 \cdot 10^{6 x}=-80 \text {. }
−
16
⋅
1
0
6
x
=
−
80
.
\newline
Solve the equation for
x
x
x
. Express the solution as a logarithm in base
10
10
10
.
\newline
x
=
x=
x
=
\newline
Approximate the value of
x
x
x
. Round your answer to the nearest thousandth.
\newline
x
≈
x \approx
x
≈
Get tutor help
Consider the equation
1
0
2
z
3
=
15
10^{\frac{2 z}{3}}=15
1
0
3
2
z
=
15
.
\newline
Solve the equation for
z
z
z
. Express the solution as a logarithm in base
10
10
10
.
\newline
z
=
z=
z
=
\newline
Approximate the value of
z
z
z
. Round your answer to the nearest thousandth.
\newline
z
≈
z \approx
z
≈
Get tutor help
Consider the equation
\newline
0.3
⋅
e
3
x
=
27
.
0.3 \cdot e^{3 x}=27 \text {. }
0.3
⋅
e
3
x
=
27
.
\newline
Solve the equation for
x
x
x
. Express the solution as a logarithm in basee.
\newline
x
=
x=
x
=
\newline
Approximate the value of
x
x
x
. Round your answer to the nearest thousandth.
\newline
x
≈
x \approx
x
≈
Get tutor help
Consider the equation
\newline
5
⋅
1
0
z
4
=
32
.
5 \cdot 10^{\frac{z}{4}}=32 \text {. }
5
⋅
1
0
4
z
=
32
.
\newline
Solve the equation for
z
z
z
. Express the solution as a logarithm in base
10
10
10
.
\newline
z
=
z=
z
=
\newline
Approximate the value of
z
z
z
. Round your answer to the nearest thousandth.
\newline
z
≈
z \approx
z
≈
Get tutor help
Let
g
(
x
)
=
8
x
−
5
g(x)=8 x-5
g
(
x
)
=
8
x
−
5
. Which of the following is equivalent to
g
(
g
(
x
)
)
g(g(x))
g
(
g
(
x
))
?
\newline
Choose
1
1
1
answer:
\newline
(A)
64
x
−
10
64 x-10
64
x
−
10
\newline
(B)
64
x
−
45
64 x-45
64
x
−
45
\newline
(C)
64
x
2
+
25
64 x^{2}+25
64
x
2
+
25
\newline
(D)
64
x
2
−
80
x
+
25
64 x^{2}-80 x+25
64
x
2
−
80
x
+
25
Get tutor help
Let
f
(
x
)
=
5
x
f(x)=5^{x}
f
(
x
)
=
5
x
and
g
(
x
)
=
x
2
+
1
g(x)=x^{2}+1
g
(
x
)
=
x
2
+
1
. What is the value of
g
(
f
(
1
)
−
2
)
g(f(1)-2)
g
(
f
(
1
)
−
2
)
?
Get tutor help
Let
f
(
x
)
=
3
4
x
+
10
f(x)=\frac{3}{4} x+10
f
(
x
)
=
4
3
x
+
10
and
g
(
x
)
=
x
2
−
3
g(x)=x^{2}-3
g
(
x
)
=
x
2
−
3
. What is the value of
f
(
−
2
−
g
(
3
)
)
f(-2-g(3))
f
(
−
2
−
g
(
3
))
?
Get tutor help
If
2
x
2
−
7
x
−
5
=
0
2 x^{2}-7 x-5=0
2
x
2
−
7
x
−
5
=
0
, what are the values of
x
x
x
?
\newline
Choose
1
1
1
answer:
\newline
(A)
1
1
1
and
5
2
\frac{5}{2}
2
5
\newline
(B)
7
4
−
89
4
\frac{7}{4}-\frac{\sqrt{89}}{4}
4
7
−
4
89
and
\newline
7
4
+
89
4
\frac{7}{4}+\frac{\sqrt{89}}{4}
4
7
+
4
89
\newline
(C)
7
2
−
69
2
\frac{7}{2}-\frac{\sqrt{69}}{2}
2
7
−
2
69
and
\newline
7
2
+
69
2
\frac{7}{2}+\frac{\sqrt{69}}{2}
2
7
+
2
69
\newline
(D)
7
2
−
89
2
\frac{7}{2}-\frac{\sqrt{89}}{2}
2
7
−
2
89
and
\newline
7
2
+
89
2
\frac{7}{2}+\frac{\sqrt{89}}{2}
2
7
+
2
89
Get tutor help
Let
f
(
x
)
=
2
x
f(x)=2^{x}
f
(
x
)
=
2
x
and
g
(
x
)
=
x
2
+
4
g(x)=x^{2}+4
g
(
x
)
=
x
2
+
4
. What is the value of
f
(
1
+
g
(
0
)
)
f(1+g(0))
f
(
1
+
g
(
0
))
?
Get tutor help
Let
f
(
x
)
=
1
x
4
+
5
f(x)=\frac{1}{x^{4}+5}
f
(
x
)
=
x
4
+
5
1
and
g
(
x
)
=
1
x
2
+
1
g(x)=\frac{1}{x^{2}+1}
g
(
x
)
=
x
2
+
1
1
. What is the value of
f
(
1
−
g
(
0
)
)
f(1-g(0))
f
(
1
−
g
(
0
))
?
Get tutor help
Let
f
(
x
)
=
2
x
+
3
f(x)=2 x+3
f
(
x
)
=
2
x
+
3
and let
g
(
x
)
=
x
2
−
4
x
g(x)=x^{2}-4 x
g
(
x
)
=
x
2
−
4
x
. Which of the following is equivalent to
g
(
f
(
x
)
)
g(f(x))
g
(
f
(
x
))
?
\newline
Choose
1
1
1
answer:
\newline
(A)
4
x
2
+
4
x
−
3
4 x^{2}+4 x-3
4
x
2
+
4
x
−
3
\newline
(B)
4
x
2
−
8
x
−
3
4 x^{2}-8 x-3
4
x
2
−
8
x
−
3
\newline
(C)
4
x
2
+
4
x
+
21
4 x^{2}+4 x+21
4
x
2
+
4
x
+
21
\newline
(D)
2
x
2
−
8
x
+
3
2 x^{2}-8 x+3
2
x
2
−
8
x
+
3
Get tutor help
Let
f
(
x
)
=
2
x
−
2
x
2
+
1
f(x)=\frac{2 x-2}{x^{2}+1}
f
(
x
)
=
x
2
+
1
2
x
−
2
and
g
(
x
)
=
x
2
+
1
g(x)=x^{2}+1
g
(
x
)
=
x
2
+
1
. What is the value of
f
(
1
+
g
(
1
)
)
f(1+g(1))
f
(
1
+
g
(
1
))
?
Get tutor help
c
=
4
b
d
3
c=\frac{4 b}{\sqrt[3]{d}}
c
=
3
d
4
b
\newline
The formula gives the capsize screening value,
c
c
c
, for a sailboat with a beam
b
b
b
feet long and that displaces
d
d
d
pounds of water. Higher capsize screening values suggest that a sailboat is more stable. Which of the following equations correctly gives the displacement in terms of the capsize screening value and the beam length?
\newline
Choose
1
1
1
answer:
\newline
(A)
d
=
(
4
b
)
3
c
d=\frac{(4 b)^{3}}{c}
d
=
c
(
4
b
)
3
\newline
(B)
d
=
c
3
4
b
d=\frac{c^{3}}{4 b}
d
=
4
b
c
3
\newline
(C)
d
=
(
4
b
c
)
3
d=\left(\frac{4 b}{c}\right)^{3}
d
=
(
c
4
b
)
3
\newline
(D)
d
=
(
c
4
b
)
3
d=\left(\frac{c}{4 b}\right)^{3}
d
=
(
4
b
c
)
3
Get tutor help
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