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Let’s check out your problem:
The graph of function
g
g
g
is shown below. Let
f
(
x
)
=
∫
0
x
g
(
t
)
d
t
f(x)=\int_{0}^{x} g(t) d t
f
(
x
)
=
∫
0
x
g
(
t
)
d
t
.
\newline
Evaluate
f
(
5
)
f(5)
f
(
5
)
.
\newline
f
(
5
)
=
f(5)=
f
(
5
)
=
View step-by-step help
Home
Math Problems
Precalculus
Evaluate functions
Full solution
Q.
The graph of function
g
g
g
is shown below. Let
f
(
x
)
=
∫
0
x
g
(
t
)
d
t
f(x)=\int_{0}^{x} g(t) d t
f
(
x
)
=
∫
0
x
g
(
t
)
d
t
.
\newline
Evaluate
f
(
5
)
f(5)
f
(
5
)
.
\newline
f
(
5
)
=
f(5)=
f
(
5
)
=
Identify Area Under Graph:
Identify the area under the graph of
g
(
t
)
g(t)
g
(
t
)
from
0
0
0
to
5
5
5
, since
f
(
x
)
f(x)
f
(
x
)
is the integral of
g
(
t
)
g(t)
g
(
t
)
from
0
0
0
to
x
x
x
,
Assume Function for
g
(
t
)
g(t)
g
(
t
)
:
Since the graph is not provided, assume
g
(
t
)
g(t)
g
(
t
)
is a function whose integral from
0
0
0
to
5
5
5
can be determined,
Calculate Integral of
g
(
t
)
g(t)
g
(
t
)
:
Calculate the integral of
g
(
t
)
g(t)
g
(
t
)
from
0
0
0
to
5
5
5
, which is
f
(
5
)
f(5)
f
(
5
)
,
Unable to Proceed Further:
Without the graph or a description of
g
(
t
)
g(t)
g
(
t
)
, we cannot proceed further,
More problems from Evaluate functions
Question
We want to factor the following expression:
\newline
x
4
+
9
x^{4}+9
x
4
+
9
\newline
Which pattern can we use to factor the expression?
\newline
U
U
U
and
V
V
V
are either constant integers or single-variable expressions.
\newline
Choose
1
1
1
answer:
\newline
(A)
(
U
+
V
)
2
(U+V)^{2}
(
U
+
V
)
2
or
(
U
−
V
)
2
(U-V)^{2}
(
U
−
V
)
2
\newline
(B)
(
U
+
V
)
(
U
−
V
)
(U+V)(U-V)
(
U
+
V
)
(
U
−
V
)
\newline
(C) We can't use any of the patterns.
Get tutor help
Posted 10 months ago
Question
z
=
−
19
+
3.14
i
z=-19+3.14 i
z
=
−
19
+
3.14
i
\newline
What are the real and imaginary parts of
z
z
z
?
\newline
Choose
1
1
1
answer:
\newline
(A)
\newline
Re
(
z
)
=
−
19
and
Im
(
z
)
=
3.14
\begin{array}{l} \operatorname{Re}(z)=-19 \text { and } \\ \operatorname{Im}(z)=3.14 \end{array}
Re
(
z
)
=
−
19
and
Im
(
z
)
=
3.14
\newline
(B)
\newline
Re
(
z
)
=
3.14
i
and
Im
(
z
)
=
−
19
\begin{array}{l} \operatorname{Re}(z)=3.14 i \text { and } \\ \operatorname{Im}(z)=-19 \end{array}
Re
(
z
)
=
3.14
i
and
Im
(
z
)
=
−
19
\newline
(c)
\newline
Re
(
z
)
=
−
19
and
Im
(
z
)
=
3.14
i
\begin{array}{l} \operatorname{Re}(z)=-19 \text { and } \\ \operatorname{Im}(z)=3.14 i \end{array}
Re
(
z
)
=
−
19
and
Im
(
z
)
=
3.14
i
\newline
(D)
\newline
Re
(
z
)
=
3.14
and
Im
(
z
)
=
−
19
\begin{array}{l} \operatorname{Re}(z)=3.14 \text { and } \\ \operatorname{Im}(z)=-19 \end{array}
Re
(
z
)
=
3.14
and
Im
(
z
)
=
−
19
Get tutor help
Posted 10 months ago
Question
z
=
−
60
−
12
i
Re
(
z
)
=
Im
(
z
)
=
\begin{array}{l}z=-60-12 i \\ \operatorname{Re}(z)= \\ \operatorname{Im}(z)=\end{array}
z
=
−
60
−
12
i
Re
(
z
)
=
Im
(
z
)
=
Get tutor help
Posted 10 months ago
Question
z
=
−
45
i
−
15.5
z=-45 i-15.5
z
=
−
45
i
−
15.5
\newline
What are the real and imaginary parts of
z
z
z
?
\newline
Choose
1
1
1
answer:
\newline
(A)
\newline
Re
(
z
)
=
−
45
and
Im
(
z
)
=
−
15.5
\begin{array}{l} \operatorname{Re}(z)=-45 \text { and } \\ \operatorname{Im}(z)=-15.5 \end{array}
Re
(
z
)
=
−
45
and
Im
(
z
)
=
−
15.5
\newline
(B)
\newline
Re
(
z
)
=
−
15.5
and
Im
(
z
)
=
−
45
\begin{array}{l} \operatorname{Re}(z)=-15.5 \text { and } \\ \operatorname{Im}(z)=-45 \end{array}
Re
(
z
)
=
−
15.5
and
Im
(
z
)
=
−
45
\newline
(C)
\newline
Re
(
z
)
=
−
45
i
and
Im
(
z
)
=
−
15.5
\begin{array}{l} \operatorname{Re}(z)=-45 i \text { and } \\ \operatorname{Im}(z)=-15.5 \end{array}
Re
(
z
)
=
−
45
i
and
Im
(
z
)
=
−
15.5
\newline
(D)
\newline
Re
(
z
)
=
−
15.5
and
Im
(
z
)
=
−
45
i
\begin{array}{l} \operatorname{Re}(z)=-15.5 \text { and } \\ \operatorname{Im}(z)=-45 i \end{array}
Re
(
z
)
=
−
15.5
and
Im
(
z
)
=
−
45
i
Get tutor help
Posted 10 months ago
Question
z
=
−
12
i
+
11
z=-12 i+11
z
=
−
12
i
+
11
\newline
What are the real and imaginary parts of
z
z
z
?
\newline
Choose
1
1
1
answer:
\newline
(A)
\newline
Re
(
z
)
=
11
and
Im
(
z
)
=
−
12
\begin{array}{l} \operatorname{Re}(z)=11 \text { and } \\ \operatorname{Im}(z)=-12 \end{array}
Re
(
z
)
=
11
and
Im
(
z
)
=
−
12
\newline
(B)
\newline
Re
(
z
)
=
11
and
Im
(
z
)
=
−
12
i
\begin{array}{l} \operatorname{Re}(z)=11 \text { and } \\ \operatorname{Im}(z)=-12 i \end{array}
Re
(
z
)
=
11
and
Im
(
z
)
=
−
12
i
\newline
(C)
\newline
Re
(
z
)
=
−
12
i
and
Im
(
z
)
=
11
\begin{array}{l} \operatorname{Re}(z)=-12 i \text { and } \\ \operatorname{Im}(z)=11 \end{array}
Re
(
z
)
=
−
12
i
and
Im
(
z
)
=
11
\newline
(D)
\newline
Re
(
z
)
=
−
12
and
Im
(
z
)
=
11
\begin{array}{l} \operatorname{Re}(z)=-12 \text { and } \\ \operatorname{Im}(z)=11 \end{array}
Re
(
z
)
=
−
12
and
Im
(
z
)
=
11
Get tutor help
Posted 10 months ago
Question
(
18
+
4
i
)
+
(
−
11
+
23
i
)
=
(18+4 i)+(-11+23 i)=
(
18
+
4
i
)
+
(
−
11
+
23
i
)
=
\newline
Express your answer in the form
(
a
+
b
i
)
(a+b i)
(
a
+
bi
)
.
Get tutor help
Posted 10 months ago
Question
(
14
+
60
i
)
+
(
−
30
+
2
i
)
=
(14+60 i)+(-30+2 i)=
(
14
+
60
i
)
+
(
−
30
+
2
i
)
=
\newline
Express your answer in the form
(
a
+
b
i
)
(a+b i)
(
a
+
bi
)
.
Get tutor help
Posted 10 months ago
Question
(
3
−
91
i
)
+
(
67
)
=
(3-91 i)+(67)=
(
3
−
91
i
)
+
(
67
)
=
\newline
Express your answer in the form
(
a
+
b
i
)
(a+b i)
(
a
+
bi
)
.
Get tutor help
Posted 10 months ago
Question
(
−
71
+
2
i
)
+
(
88
−
12
i
)
=
(-71+2 i)+(88-12 i)=
(
−
71
+
2
i
)
+
(
88
−
12
i
)
=
\newline
Express your answer in the form
(
a
+
b
i
)
(a+b i)
(
a
+
bi
)
.
Get tutor help
Posted 10 months ago
Question
(
−
33
−
2
i
)
−
(
50
+
9
i
)
=
(-33-2 i)-(50+9 i)=
(
−
33
−
2
i
)
−
(
50
+
9
i
)
=
\newline
Express your answer in the form
(
a
+
b
i
)
(a+b i)
(
a
+
bi
)
.
Get tutor help
Posted 10 months ago
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