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What Is the coefficient of the second term in this expression?
5
m
+
k
+
3
n
5m+k+3n
5
m
+
k
+
3
n
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Math Problems
Calculus
Find second derivatives of trigonometric, exponential, and logarithmic functions
Full solution
Q.
What Is the coefficient of the second term in this expression?
5
m
+
k
+
3
n
5m+k+3n
5
m
+
k
+
3
n
Identify Second Term:
Identify the second term in the expression
5
m
+
k
+
3
n
5m+k+3n
5
m
+
k
+
3
n
. The second term is
k
k
k
.
Determine Coefficient:
Determine the coefficient of the term
k
k
k
. Since there's no number written in front of
k
k
k
, the coefficient is
1
1
1
.
More problems from Find second derivatives of trigonometric, exponential, and logarithmic functions
Question
f
′
(
x
)
=
12
e
x
and
f
(
4
)
=
−
16
+
12
e
4
.
f
(
0
)
=
\begin{array}{l}f^{\prime}(x)=12 e^{x} \text { and } f(4)=-16+12 e^{4} . \\ f(0)=\end{array}
f
′
(
x
)
=
12
e
x
and
f
(
4
)
=
−
16
+
12
e
4
.
f
(
0
)
=
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Posted 1 month ago
Question
Solve for
n
n
n
.
8
3
=
12
n
n
=
□
\begin{array}{l}\frac{8}{3}=\frac{12}{n} \\ n=\square\end{array}
3
8
=
n
12
n
=
□
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Posted 2 months ago
Question
Solve the equation.
b
6
=
3
b
=
□
\begin{align*} \frac{b}{6} &= 3 \\ b &= \Box \end{align*}
6
b
b
=
3
=
□
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Posted 2 months ago
Question
Solve the equation. \begin{align*} \frac{b}{6} &= 3, \ b &= \Box \end{align*}
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Posted 2 months ago
Question
Solve for
n
n
n
.
12
5
=
n
8
\frac{12}{5}=\frac{n}{8}
5
12
=
8
n
\newline
n
=
□
n = \Box
n
=
□
Get tutor help
Posted 2 months ago
Question
The second derivative of the function
f
f
f
is defined by
f
′
′
(
x
)
=
x
2
−
5
cos
(
2
x
)
f^{\prime \prime}(x)=x^{2}-5 \cos (2 x)
f
′′
(
x
)
=
x
2
−
5
cos
(
2
x
)
for
−
2.5
<
x
<
3.5
-2.5<x<3.5
−
2.5
<
x
<
3.5
. Find the
x
x
x
-values, if any, in the given domain where the function
f
f
f
has an inflection point. You may use a calculator and round all values to
3
3
3
decimal places.
\newline
Answer:
x
=
x=
x
=
Get tutor help
Posted 2 months ago
Question
f
′
(
x
)
=
12
e
x
and
f
(
4
)
=
−
16
+
12
e
4
.
f
(
0
)
=
\begin{array}{l}f^{\prime}(x)=12 e^{x} \text { and } f(4)=-16+12 e^{4} . \\ f(0)=\end{array}
f
′
(
x
)
=
12
e
x
and
f
(
4
)
=
−
16
+
12
e
4
.
f
(
0
)
=
Get tutor help
Posted 2 months ago
Question
x
2
−
6
x
+
11
=
y
x
=
y
+
1
\begin{array}{c}x^{2}-6 x+11=y \\ x=y+1\end{array}
x
2
−
6
x
+
11
=
y
x
=
y
+
1
Get tutor help
Posted 2 months ago
Question
=
x
2
+
4
=
(
g
+
h
)
(
x
)
\begin{array}{l} =x^{2}+4 \\ =(g+h)(x) \end{array}
=
x
2
+
4
=
(
g
+
h
)
(
x
)
\newline
9
9
9
.
\newline
h
(
x
)
=
x
2
+
x
g
(
x
)
=
3
x
+
5
Find
(
h
⋅
g
)
(
x
)
\begin{array}{l} h(x)=x^{2}+x \\ g(x)=3 x+5 \\ \text { Find }(h \cdot g)(x) \end{array}
h
(
x
)
=
x
2
+
x
g
(
x
)
=
3
x
+
5
Find
(
h
⋅
g
)
(
x
)
Get tutor help
Posted 2 months ago
Question
Factor completely:
\newline
8
c
3
+
27
m
6
8 c^{3}+27 m^{6}
8
c
3
+
27
m
6
\newline
Answer:
Get tutor help
Posted 2 months ago