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Math Problems
Calculus
Find derivatives of sine and cosine functions
Determine whether the function
f
(
x
)
f(x)
f
(
x
)
is continuous at
x
=
5
x=5
x
=
5
.
\newline
f
(
x
)
=
{
12
−
x
2
,
x
<
5
−
9
−
x
,
x
≥
5
f(x)=\left\{\begin{array}{ll} 12-x^{2}, & x<5 \\ -9-x, & x \geq 5 \end{array}\right.
f
(
x
)
=
{
12
−
x
2
,
−
9
−
x
,
x
<
5
x
≥
5
\newline
f
(
x
)
f(x)
f
(
x
)
is continuous at
x
=
5
x=5
x
=
5
\newline
f
(
x
)
f(x)
f
(
x
)
is discontinuous at
x
=
5
x=5
x
=
5
Get tutor help
Determine whether the function
f
(
x
)
f(x)
f
(
x
)
is continuous at
x
=
3
x=3
x
=
3
.
\newline
f
(
x
)
=
{
7
−
2
x
2
,
x
≤
3
−
8
−
x
,
x
>
3
f(x)=\left\{\begin{array}{ll} 7-2 x^{2}, & x \leq 3 \\ -8-x, & x>3 \end{array}\right.
f
(
x
)
=
{
7
−
2
x
2
,
−
8
−
x
,
x
≤
3
x
>
3
\newline
f
(
x
)
f(x)
f
(
x
)
is continuous at
x
=
3
x=3
x
=
3
\newline
Submit Answer
\newline
f
(
x
)
f(x)
f
(
x
)
is discontinuous at
x
=
3
x=3
x
=
3
Get tutor help
Determine whether the function
f
(
x
)
f(x)
f
(
x
)
is continuous at
x
=
3
x=3
x
=
3
.
\newline
f
(
x
)
=
{
4
−
2
x
2
,
x
>
3
−
11
−
2
x
,
x
≤
3
f(x)=\left\{\begin{array}{ll} 4-2 x^{2}, & x>3 \\ -11-2 x, & x \leq 3 \end{array}\right.
f
(
x
)
=
{
4
−
2
x
2
,
−
11
−
2
x
,
x
>
3
x
≤
3
\newline
f
(
x
)
f(x)
f
(
x
)
is discontinuous at
x
=
3
x=3
x
=
3
\newline
f
(
x
)
f(x)
f
(
x
)
is continuous at
x
=
3
x=3
x
=
3
Get tutor help
Determine whether the function
f
(
x
)
f(x)
f
(
x
)
is continuous at
x
=
3
x=3
x
=
3
.
\newline
f
(
x
)
=
{
10
−
3
x
2
,
x
<
−
3
−
5
+
3
x
,
x
≥
−
3
f(x)=\left\{\begin{array}{ll} 10-3 x^{2}, & x<-3 \\ -5+3 x, & x \geq-3 \end{array}\right.
f
(
x
)
=
{
10
−
3
x
2
,
−
5
+
3
x
,
x
<
−
3
x
≥
−
3
\newline
f
(
x
)
f(x)
f
(
x
)
is discontinuous at
x
=
3
x=3
x
=
3
\newline
f
(
x
)
f(x)
f
(
x
)
is continuous at
x
=
3
x=3
x
=
3
Get tutor help
Determine whether the function
f
(
x
)
f(x)
f
(
x
)
is continuous at
x
=
2
x=2
x
=
2
.
\newline
f
(
x
)
=
{
2
−
x
2
,
x
<
2
−
8
+
3
x
,
x
≥
2
f(x)=\left\{\begin{array}{ll} 2-x^{2}, & x<2 \\ -8+3 x, & x \geq 2 \end{array}\right.
f
(
x
)
=
{
2
−
x
2
,
−
8
+
3
x
,
x
<
2
x
≥
2
\newline
f
(
x
)
f(x)
f
(
x
)
is continuous at
x
=
2
x=2
x
=
2
\newline
Submit Answer
\newline
f
(
x
)
f(x)
f
(
x
)
is discontinuous at
x
=
2
x=2
x
=
2
Get tutor help
Determine whether the function
f
(
x
)
f(x)
f
(
x
)
is continuous at
x
=
5
x=5
x
=
5
.
\newline
f
(
x
)
=
{
12
−
x
2
,
x
<
5
−
9
−
x
,
x
≥
5
f(x)=\left\{\begin{array}{ll} 12-x^{2}, & x<5 \\ -9-x, & x \geq 5 \end{array}\right.
f
(
x
)
=
{
12
−
x
2
,
−
9
−
x
,
x
<
5
x
≥
5
\newline
f
(
x
)
f(x)
f
(
x
)
is continuous at
x
=
5
x=5
x
=
5
\newline
Submit Answer
\newline
f
(
x
)
f(x)
f
(
x
)
is discontinuous at
x
=
5
x=5
x
=
5
Get tutor help
Determine whether the function
f
(
x
)
f(x)
f
(
x
)
is continuous at
x
=
5
x=5
x
=
5
.
\newline
f
(
x
)
=
{
3
+
x
2
,
x
≤
5
18
+
2
x
,
x
>
5
f(x)=\left\{\begin{array}{ll} 3+x^{2}, & x \leq 5 \\ 18+2 x, & x>5 \end{array}\right.
f
(
x
)
=
{
3
+
x
2
,
18
+
2
x
,
x
≤
5
x
>
5
\newline
f
(
x
)
f(x)
f
(
x
)
is continuous at
x
=
5
x=5
x
=
5
\newline
f
(
x
)
f(x)
f
(
x
)
is discontinuous at
x
=
5
x=5
x
=
5
Get tutor help
Determine whether the function
f
(
x
)
f(x)
f
(
x
)
is continuous at
x
=
−
4
x=-4
x
=
−
4
.
\newline
f
(
x
)
=
{
1
+
x
2
,
x
≥
4
7
+
2
x
,
x
<
4
f(x)=\left\{\begin{array}{ll} 1+x^{2}, & x \geq 4 \\ 7+2 x, & x<4 \end{array}\right.
f
(
x
)
=
{
1
+
x
2
,
7
+
2
x
,
x
≥
4
x
<
4
\newline
f
(
x
)
f(x)
f
(
x
)
is discontinuous at
x
=
−
4
x=-4
x
=
−
4
\newline
f
(
x
)
f(x)
f
(
x
)
is continuous at
x
=
−
4
x=-4
x
=
−
4
Get tutor help
Determine whether the function
f
(
x
)
f(x)
f
(
x
)
is continuous at
x
=
−
2
x=-2
x
=
−
2
.
\newline
f
(
x
)
=
{
1
−
x
2
,
x
≥
−
2
−
9
−
3
x
,
x
<
−
2
f(x)=\left\{\begin{array}{ll} 1-x^{2}, & x \geq-2 \\ -9-3 x, & x<-2 \end{array}\right.
f
(
x
)
=
{
1
−
x
2
,
−
9
−
3
x
,
x
≥
−
2
x
<
−
2
\newline
f
(
x
)
f(x)
f
(
x
)
is discontinuous at
x
=
−
2
x=-2
x
=
−
2
\newline
f
(
x
)
f(x)
f
(
x
)
is continuous at
x
=
−
2
x=-2
x
=
−
2
Get tutor help
Find
(
f
∘
g
)
(
0
)
(f \circ g)(0)
(
f
∘
g
)
(
0
)
.
f
(
x
)
=
x
+
6
f(x) = x + 6
f
(
x
)
=
x
+
6
,
g
(
x
)
=
4
x
g(x) = 4x
g
(
x
)
=
4
x
Get tutor help
h
(
n
)
=
−
10
+
12
n
h(n)=-10+12n
h
(
n
)
=
−
10
+
12
n
\newline
Complete the recursive formula of
h
(
n
)
h(n)
h
(
n
)
.
\newline
h
(
1
)
=
□
h(1)=\square
h
(
1
)
=
□
\newline
h
(
n
)
=
h
(
n
−
1
)
+
□
h(n)=h(n-1)+\square
h
(
n
)
=
h
(
n
−
1
)
+
□
Get tutor help
f
(
n
)
=
41
−
5
n
f(n)=41-5n
f
(
n
)
=
41
−
5
n
\newline
Complete the recursive formula of
f
(
n
)
f(n)
f
(
n
)
.
\newline
f
(
1
)
=
□
f(1)=\square
f
(
1
)
=
□
\newline
f
(
n
)
=
f
(
n
−
1
)
+
□
f(n)=f(n-1)+\square
f
(
n
)
=
f
(
n
−
1
)
+
□
Get tutor help
Solve the equation.
6
=
k
9
6=\frac{k}{9}
6
=
9
k
,
k
=
□
k=\square
k
=
□
Get tutor help
Solve the equation.
\newline
6
=
k
9
6=\frac{k}{9}
6
=
9
k
\newline
k
=
□
k=\square
k
=
□
Get tutor help
Solve the equation. \begin{align*} 6 &= \frac{x}{5} \\ x &= \square \end{align*}
Get tutor help
The rate of change
d
P
d
t
\frac{d P}{d t}
d
t
d
P
of the number of people on an island is modeled by the following differential equation:
\newline
d
P
d
t
=
3920
27101
P
(
1
−
P
784
)
\frac{d P}{d t}=\frac{3920}{27101} P\left(1-\frac{P}{784}\right)
d
t
d
P
=
27101
3920
P
(
1
−
784
P
)
\newline
At
t
=
0
t=0
t
=
0
, the number of people on the island is
123
123
123
and is increasing at a rate of
15
15
15
people per hour. Find
lim
t
→
∞
P
(
t
)
\lim _{t \rightarrow \infty} P(t)
lim
t
→
∞
P
(
t
)
.
\newline
Answer:
Get tutor help
The rate of change
d
P
d
t
\frac{d P}{d t}
d
t
d
P
of the number of people in a park is modeled by the following differential equation:
\newline
d
P
d
t
=
4085
36816
P
(
1
−
P
860
)
\frac{d P}{d t}=\frac{4085}{36816} P\left(1-\frac{P}{860}\right)
d
t
d
P
=
36816
4085
P
(
1
−
860
P
)
\newline
At
t
=
0
t=0
t
=
0
, the number of people in the park is
236
236
236
and is increasing at a rate of
19
19
19
people per hour. Find
lim
t
→
∞
P
′
(
t
)
\lim _{t \rightarrow \infty} P^{\prime}(t)
lim
t
→
∞
P
′
(
t
)
.
\newline
Answer:
Get tutor help
Find
(
f
∘
g
)
(
0
)
(f \circ g)(0)
(
f
∘
g
)
(
0
)
.
\newline
\begin{align*} &f(x) = x + 5,\ &g(x) = x^{2} + 3,\ &(f \circ g)(0) = \end{align*}
Get tutor help
If
lim
x
→
−
6
f
(
x
)
=
9
\lim_{x \to -6}f(x)=9
lim
x
→
−
6
f
(
x
)
=
9
and
lim
x
→
−
6
k
(
x
)
=
10
\lim_{x \to -6}k(x)=10
lim
x
→
−
6
k
(
x
)
=
10
, what is the value of
lim
x
→
−
6
[
k
(
x
)
−
f
(
x
)
]
\lim_{x \to -6}[k(x)-f(x)]
lim
x
→
−
6
[
k
(
x
)
−
f
(
x
)]
?
Get tutor help
A particle moves along the
x
x
x
-axis such that at any time
t
≥
0
t \geq 0
t
≥
0
its position is
x
(
t
)
x(t)
x
(
t
)
, its velocity is
v
(
t
)
v(t)
v
(
t
)
, and its acceleration is
a
(
t
)
a(t)
a
(
t
)
.
\newline
What is the average velocity of the particle on the interval
1
≤
t
≤
8
1 \leq t \leq 8
1
≤
t
≤
8
?
\newline
v
(
8
)
−
v
(
1
)
7
\frac{v(8)-v(1)}{7}
7
v
(
8
)
−
v
(
1
)
\newline
a
(
8
)
−
a
(
1
)
7
\frac{a(8)-a(1)}{7}
7
a
(
8
)
−
a
(
1
)
\newline
1
7
∫
1
8
a
(
t
)
d
t
\frac{1}{7} \int_{1}^{8} a(t) d t
7
1
∫
1
8
a
(
t
)
d
t
\newline
1
7
∫
1
8
v
(
t
)
d
t
\frac{1}{7} \int_{1}^{8} v(t) d t
7
1
∫
1
8
v
(
t
)
d
t
Get tutor help
A particle moves along the
x
x
x
-axis such that at any time
t
≥
0
t \geq 0
t
≥
0
its position is
x
(
t
)
x(t)
x
(
t
)
, its velocity is
v
(
t
)
v(t)
v
(
t
)
, and its acceleration is
a
(
t
)
a(t)
a
(
t
)
.
\newline
What is the average velocity of the particle on the interval
0
≤
t
≤
8
0 \leq t \leq 8
0
≤
t
≤
8
?
\newline
v
(
8
)
−
v
(
0
)
8
\frac{v(8)-v(0)}{8}
8
v
(
8
)
−
v
(
0
)
\newline
1
8
∫
0
8
a
(
t
)
d
t
\frac{1}{8} \int_{0}^{8} a(t) d t
8
1
∫
0
8
a
(
t
)
d
t
\newline
x
(
8
)
−
x
(
0
)
8
\frac{x(8)-x(0)}{8}
8
x
(
8
)
−
x
(
0
)
\newline
∫
0
8
v
(
t
)
d
t
\int_{0}^{8} v(t) d t
∫
0
8
v
(
t
)
d
t
Get tutor help
A particle moves along the
x
x
x
-axis such that at any time
t
≥
0
t \geq 0
t
≥
0
its position is
x
(
t
)
x(t)
x
(
t
)
, its velocity is
v
(
t
)
v(t)
v
(
t
)
, and its acceleration is
a
(
t
)
a(t)
a
(
t
)
.
\newline
What is the average velocity of the particle on the interval
0
≤
t
≤
8
0 \leq t \leq 8
0
≤
t
≤
8
?
\newline
v
(
8
)
−
v
(
0
)
8
\frac{v(8)-v(0)}{8}
8
v
(
8
)
−
v
(
0
)
\newline
a
(
8
)
−
a
(
0
)
8
\frac{a(8)-a(0)}{8}
8
a
(
8
)
−
a
(
0
)
\newline
1
8
∫
0
8
v
(
t
)
d
t
\frac{1}{8} \int_{0}^{8} v(t) d t
8
1
∫
0
8
v
(
t
)
d
t
\newline
1
8
∫
0
8
a
(
t
)
d
t
\frac{1}{8} \int_{0}^{8} a(t) d t
8
1
∫
0
8
a
(
t
)
d
t
Get tutor help
A particle moves along the
x
x
x
-axis such that at any time
t
≥
0
t \geq 0
t
≥
0
its position is
x
(
t
)
x(t)
x
(
t
)
, its velocity is
v
(
t
)
v(t)
v
(
t
)
, and its acceleration is
a
(
t
)
a(t)
a
(
t
)
.
\newline
What is the average acceleration of the particle on the interval
0
≤
t
≤
8
?
0 \leq t \leq 8 ?
0
≤
t
≤
8
?
\newline
x
(
8
)
−
x
(
0
)
8
\frac{x(8)-x(0)}{8}
8
x
(
8
)
−
x
(
0
)
\newline
a
(
8
)
−
a
(
0
)
8
\frac{a(8)-a(0)}{8}
8
a
(
8
)
−
a
(
0
)
\newline
1
8
∫
0
8
a
(
t
)
d
t
\frac{1}{8} \int_{0}^{8} a(t) d t
8
1
∫
0
8
a
(
t
)
d
t
\newline
1
8
∫
0
8
v
(
t
)
d
t
\frac{1}{8} \int_{0}^{8} v(t) d t
8
1
∫
0
8
v
(
t
)
d
t
Get tutor help
The rate at which the amount of water in a tank is changing can be measured by the differentiable function
f
f
f
, where
f
(
t
)
f(t)
f
(
t
)
is measured in liters per second and
t
t
t
is measured in seconds. What are the units of
∫
3
5
f
(
t
)
d
t
\int_{3}^{5} f(t) d t
∫
3
5
f
(
t
)
d
t
?
\newline
liters
\newline
seconds
\newline
liters / second
\newline
seconds / liter
\newline
liters
/
/
/
second
2
^{2}
2
\newline
seconds / liter
2
{ }^{2}
2
Get tutor help
The atmospheric pressure of the air changes with height above sea level. The pressure of the air at a given height above sea level can be measured by the differentiable function
f
(
h
)
f(h)
f
(
h
)
, in psi, where
h
h
h
is measured in meters. What are the units of
f
′
′
(
h
)
?
f^{\prime \prime}(h) ?
f
′′
(
h
)?
\newline
meters
\newline
psi
\newline
meters / psi
\newline
psi / meter
\newline
meters
/
p
s
i
2
/ \mathrm{psi}^{2}
/
psi
2
\newline
psi
/
/
/
meter
2
^{2}
2
Get tutor help
The number of people waiting in line to buy a new piece of technology is measured by the differentiable function
f
f
f
, where
f
(
t
)
f(t)
f
(
t
)
is measured in people and
t
t
t
is measured in minutes after the store opened. What are the units of
f
′
(
t
)
f^{\prime}(t)
f
′
(
t
)
?
\newline
minutes
\newline
people
\newline
minutes / person
\newline
people / minute
\newline
minutes / person
2
{ }^{2}
2
\newline
people / minute
2
{ }^{2}
2
Get tutor help
The atmospheric pressure of the air changes with height above sea level. The height above sea level at a given pressure can be measured by the differentiable function
f
(
p
)
f(p)
f
(
p
)
in feet, where
p
p
p
is measured in psi. What are the units of
f
′
′
(
p
)
?
f^{\prime \prime}(p) ?
f
′′
(
p
)?
\newline
feet
\newline
psi
\newline
feet / psi
\newline
psi / foot
\newline
feet
/
p
s
i
2
/ \mathrm{psi}^{2}
/
psi
2
\newline
psi / foot
2
{ }^{2}
2
Get tutor help
The amount of water in a tank is measured by the differentiable function
f
f
f
, where
f
(
t
)
f(t)
f
(
t
)
is measured in liters and
t
t
t
is measured in seconds. What are the units of
∫
0
10
f
′
(
t
)
d
t
?
\int_{0}^{10} f^{\prime}(t) d t ?
∫
0
10
f
′
(
t
)
d
t
?
\newline
liters
\newline
seconds
\newline
liters / second
\newline
seconds / liter
\newline
liters
/
/
/
second
2
^{2}
2
\newline
seconds
/
/
/
liter
2
^{2}
2
Get tutor help
The number of people waiting in line to buy a new piece of technology is measured by the differentiable function
f
f
f
, where
f
(
t
)
f(t)
f
(
t
)
is measured in people and
t
t
t
is measured in hours after the store opened. What are the units of
1
2
∫
3
5
f
(
t
)
d
t
\frac{1}{2} \int_{3}^{5} f(t) d t
2
1
∫
3
5
f
(
t
)
d
t
?
\newline
people
\newline
hours
\newline
people / hour
\newline
hours / person
\newline
people / hour
2
{ }^{2}
2
\newline
hours / person
2
{ }^{2}
2
Get tutor help
The atmospheric pressure of the air changes with height above sea level. The pressure of the air at a given height above sea level can be measured by the differentiable function
f
(
h
)
f(h)
f
(
h
)
, in psi, where
h
h
h
is measured in meters. What are the units of
f
′
(
h
)
f^{\prime}(h)
f
′
(
h
)
?
\newline
psi
\newline
meters
\newline
psi / meter
\newline
meters / psi
\newline
psi
/
/
/
meter
2
^{2}
2
\newline
meters
/
p
s
i
2
/ \mathrm{psi}^{2}
/
psi
2
Get tutor help
The rate at which the amount of water in a tank is changing can be measured by the differentiable function
f
f
f
, where
f
(
t
)
f(t)
f
(
t
)
is measured in liters per second and
t
t
t
is measured in seconds. What are the units of
∫
3
10
f
′
(
t
)
d
t
\int_{3}^{10} f^{\prime}(t) d t
∫
3
10
f
′
(
t
)
d
t
?
\newline
liters
\newline
seconds
\newline
liters / second
\newline
seconds / liter
\newline
liters
/
/
/
second
2
^{2}
2
\newline
seconds / liter
2
{ }^{2}
2
Get tutor help
The number of people waiting in line to buy a new piece of technology is measured by the differentiable function
f
f
f
, where
f
(
t
)
f(t)
f
(
t
)
is measured in people and
t
t
t
is measured in hours after the store opened. What are the units of
∫
1
5
f
′
(
t
)
d
t
\int_{1}^{5} f^{\prime}(t) d t
∫
1
5
f
′
(
t
)
d
t
?
\newline
people
\newline
hours
\newline
people / hour
\newline
hours / person
\newline
people / hour
2
{ }^{2}
2
\newline
hours / person
2
{ }^{2}
2
Get tutor help
The atmospheric pressure of the air changes with height above sea level. The height above sea level at a given pressure can be measured by the differentiable function
f
(
p
)
f(p)
f
(
p
)
in miles, where
p
p
p
is measured in psi. What are the units of
f
′
′
(
p
)
f^{\prime \prime}(p)
f
′′
(
p
)
?
\newline
psi
\newline
miles
\newline
psi / mile
\newline
miles / psi
\newline
psi
/
/
/
mile
2
^{2}
2
\newline
miles
/
p
s
i
2
/ \mathrm{psi}^{2}
/
psi
2
Get tutor help
The atmospheric pressure of the air changes with height above sea level. The height above sea level at a given pressure can be measured by the differentiable function
f
(
p
)
f(p)
f
(
p
)
in feet, where
p
p
p
is measured in psi. What are the units of
f
′
(
p
)
?
f^{\prime}(p) ?
f
′
(
p
)?
\newline
psi
\newline
feet
\newline
psi / foot
\newline
feet / psi
\newline
psi / foot
2
{ }^{2}
2
\newline
feet
/
p
s
i
2
/ \mathrm{psi}^{2}
/
psi
2
Get tutor help
Find the derivative of the following function.
\newline
y
=
6
9
x
6
+
8
x
5
y=6^{9 x^{6}+8 x^{5}}
y
=
6
9
x
6
+
8
x
5
\newline
Answer:
y
′
=
y^{\prime}=
y
′
=
Get tutor help
g
(
x
)
=
x
5
g
′
(
x
)
=
\begin{array}{l}g(x)=x^{5} \\ g^{\prime}(x)=\end{array}
g
(
x
)
=
x
5
g
′
(
x
)
=
Get tutor help
f
′
(
x
)
=
−
3
e
x
and
f
(
1
)
=
12
−
3
e
.
f
(
0
)
=
□
\begin{array}{l}f^{\prime}(x)=-3 e^{x} \text { and } f(1)=12-3 e . \\ f(0)=\square\end{array}
f
′
(
x
)
=
−
3
e
x
and
f
(
1
)
=
12
−
3
e
.
f
(
0
)
=
□
Get tutor help
f
′
(
x
)
=
3
x
2
−
2
x
+
7
and
f
(
6
)
=
200.
f
(
1
)
=
\begin{array}{l}f^{\prime}(x)=3 x^{2}-2 x+7 \text { and } f(6)=200 . \\ f(1)=\end{array}
f
′
(
x
)
=
3
x
2
−
2
x
+
7
and
f
(
6
)
=
200.
f
(
1
)
=
Get tutor help
The rate of change of the perceived stimulus
p
p
p
with respect to the measured intensity
s
s
s
of the stimulus is inversely proportional to the intensity of the stimulus.
\newline
Which equation describes this relationship?
\newline
Choose
1
1
1
answer:
\newline
(A)
d
s
d
p
=
k
s
\frac{d s}{d p}=\frac{k}{s}
d
p
d
s
=
s
k
\newline
(B)
d
p
d
s
=
k
s
\frac{d p}{d s}=\frac{k}{s}
d
s
d
p
=
s
k
\newline
(C)
d
p
d
s
=
k
p
\frac{d p}{d s}=\frac{k}{p}
d
s
d
p
=
p
k
\newline
(D)
d
s
d
p
=
k
p
\frac{d s}{d p}=\frac{k}{p}
d
p
d
s
=
p
k
Get tutor help
A particle moves along a straight line. Its speed is inversely proportional to the square of the distance,
S
S
S
, it has traveled.
\newline
Which equation describes this relationship?
\newline
Choose
1
1
1
answer:
\newline
(A)
S
(
t
)
=
k
t
2
S(t)=\frac{k}{t^{2}}
S
(
t
)
=
t
2
k
\newline
(B)
S
(
t
)
=
k
S
2
S(t)=\frac{k}{S^{2}}
S
(
t
)
=
S
2
k
\newline
(C)
d
S
d
t
=
k
t
2
\frac{d S}{d t}=\frac{k}{t^{2}}
d
t
d
S
=
t
2
k
\newline
(D)
d
S
d
t
=
k
S
2
\frac{d S}{d t}=\frac{k}{S^{2}}
d
t
d
S
=
S
2
k
Get tutor help
The warming or cooling rate of a drink is proportional to the difference between the ambient temperature
T
a
T_{a}
T
a
and the current temperature
T
T
T
of the drink.
\newline
Which equation describes this relationship?
\newline
Choose
1
1
1
answer:
\newline
(A)
d
T
d
t
=
k
(
T
a
−
T
)
\frac{d T}{d t}=\frac{k}{\left(T_{a}-T\right)}
d
t
d
T
=
(
T
a
−
T
)
k
\newline
(B)
d
T
d
t
=
k
(
T
a
−
T
)
\frac{d T}{d t}=k\left(T_{a}-T\right)
d
t
d
T
=
k
(
T
a
−
T
)
\newline
(C)
T
(
t
)
=
k
(
T
a
−
T
)
T(t)=k\left(T_{a}-T\right)
T
(
t
)
=
k
(
T
a
−
T
)
\newline
(D)
T
(
t
)
=
k
(
T
a
−
T
)
T(t)=\frac{k}{\left(T_{a}-T\right)}
T
(
t
)
=
(
T
a
−
T
)
k
Get tutor help
A radioactive material decays at a rate of change proportional to the current amount,
Q
Q
Q
, of the radioactive material.
\newline
Which equation describes this relationship?
\newline
Choose
1
1
1
answer:
\newline
(A)
Q
(
t
)
=
−
Q
k
t
Q(t)=-Q^{k t}
Q
(
t
)
=
−
Q
k
t
\newline
(B)
Q
(
t
)
=
−
k
Q
Q(t)=-k Q
Q
(
t
)
=
−
k
Q
\newline
(C)
d
Q
d
t
=
−
k
Q
\frac{d Q}{d t}=-k Q
d
t
d
Q
=
−
k
Q
\newline
(D)
d
Q
d
t
=
−
Q
k
t
\frac{d Q}{d t}=-Q^{k t}
d
t
d
Q
=
−
Q
k
t
Get tutor help
The learning rate for new skills is proportional to the difference between the maximum potential for learning that skill,
M
M
M
, and the amount of the skill already learned,
L
L
L
.
\newline
Which equation describes this relationship?
\newline
Choose
1
1
1
answer:
\newline
(A)
L
(
t
)
=
k
(
M
−
L
)
L(t)=\frac{k}{(M-L)}
L
(
t
)
=
(
M
−
L
)
k
\newline
(B)
d
L
d
t
=
k
(
M
−
L
)
\frac{d L}{d t}=k(M-L)
d
t
d
L
=
k
(
M
−
L
)
\newline
(C)
L
(
t
)
=
k
(
M
−
L
)
L(t)=k(M-L)
L
(
t
)
=
k
(
M
−
L
)
\newline
(D)
d
L
d
t
=
k
(
M
−
L
)
\frac{d L}{d t}=\frac{k}{(M-L)}
d
t
d
L
=
(
M
−
L
)
k
Get tutor help
The rate of change of the perceived stimulus
p
p
p
with respect to the measured intensity
s
s
s
of the stimulus is inversely proportional to the intensity of the stimulus.
\newline
Which equation describes this relationship?
\newline
Choose
1
1
1
answer:
\newline
(A)
d
p
d
s
=
k
p
\frac{d p}{d s}=\frac{k}{p}
d
s
d
p
=
p
k
\newline
(B)
d
p
d
s
=
k
s
\frac{d p}{d s}=\frac{k}{s}
d
s
d
p
=
s
k
\newline
(C)
d
s
d
p
=
k
p
\frac{d s}{d p}=\frac{k}{p}
d
p
d
s
=
p
k
\newline
(D)
d
s
d
p
=
k
s
\frac{d s}{d p}=\frac{k}{s}
d
p
d
s
=
s
k
Get tutor help
The warming or cooling rate of a drink is proportional to the difference between the ambient temperature
T
a
T_{a}
T
a
and the current temperature
T
T
T
of the drink.
\newline
Which equation describes this relationship?
\newline
Choose
1
1
1
answer:
\newline
(A)
T
(
t
)
=
k
(
T
a
−
T
)
T(t)=k\left(T_{a}-T\right)
T
(
t
)
=
k
(
T
a
−
T
)
\newline
(B)
d
T
d
t
=
k
(
T
a
−
T
)
\frac{d T}{d t}=k\left(T_{a}-T\right)
d
t
d
T
=
k
(
T
a
−
T
)
\newline
(C)
d
T
d
t
=
k
(
T
a
−
T
)
\frac{d T}{d t}=\frac{k}{\left(T_{a}-T\right)}
d
t
d
T
=
(
T
a
−
T
)
k
\newline
(D)
T
(
t
)
=
k
(
T
a
−
T
)
T(t)=\frac{k}{\left(T_{a}-T\right)}
T
(
t
)
=
(
T
a
−
T
)
k
Get tutor help
A radioactive material decays at a rate of change proportional to the current amount,
Q
Q
Q
, of the radioactive material.
\newline
Which equation describes this relationship?
\newline
Choose
1
1
1
answer:
\newline
(A)
d
Q
d
t
=
−
k
Q
\frac{d Q}{d t}=-k Q
d
t
d
Q
=
−
k
Q
\newline
(B)
Q
(
t
)
=
−
Q
k
t
Q(t)=-Q^{k t}
Q
(
t
)
=
−
Q
k
t
\newline
(C)
d
Q
d
t
=
−
Q
k
t
\frac{d Q}{d t}=-Q^{k t}
d
t
d
Q
=
−
Q
k
t
\newline
(D)
Q
(
t
)
=
−
k
Q
Q(t)=-k Q
Q
(
t
)
=
−
k
Q
Get tutor help
- Let
f
f
f
be a function such that
f
(
1
)
=
0
f(1)=0
f
(
1
)
=
0
and
f
′
(
1
)
=
−
7
f^{\prime}(1)=-7
f
′
(
1
)
=
−
7
.
\newline
- Let
g
g
g
be the function
g
(
x
)
=
x
g(x)=\sqrt{x}
g
(
x
)
=
x
.
\newline
Evaluate
d
d
x
[
f
(
x
)
g
(
x
)
]
\frac{d}{d x}\left[\frac{f(x)}{g(x)}\right]
d
x
d
[
g
(
x
)
f
(
x
)
]
at
x
=
1
x=1
x
=
1
.
Get tutor help
The differentiable functions
x
x
x
and
y
y
y
are related by the following equation:
\newline
sin
(
x
)
+
cos
(
y
)
=
2
\sin (x)+\cos (y)=\sqrt{2}
sin
(
x
)
+
cos
(
y
)
=
2
\newline
Also,
d
x
d
t
=
5
\frac{d x}{d t}=5
d
t
d
x
=
5
.
\newline
Find
d
y
d
t
\frac{d y}{d t}
d
t
d
y
when
y
=
π
4
y=\frac{\pi}{4}
y
=
4
π
and
0
<
x
<
π
2
0<x<\frac{\pi}{2}
0
<
x
<
2
π
.
Get tutor help
Tom was given this problem:
\newline
The side
s
(
t
)
s(t)
s
(
t
)
of a square is decreasing at a rate of
2
2
2
kilometers per hour. At a certain instant
t
0
t_{0}
t
0
, the side is
9
9
9
kilometers. What is the rate of change of the area
A
(
t
)
A(t)
A
(
t
)
of the square at that instant?
\newline
Which equation should Tom use to solve the problem?
\newline
Choose
1
1
1
answer:
\newline
(A)
A
(
t
)
=
4
⋅
s
(
t
)
A(t)=4 \cdot s(t)
A
(
t
)
=
4
⋅
s
(
t
)
\newline
(B)
A
(
t
)
=
[
s
(
t
)
]
3
A(t)=[s(t)]^{3}
A
(
t
)
=
[
s
(
t
)
]
3
\newline
(C)
A
(
t
)
=
[
s
(
t
)
]
2
A(t)=[s(t)]^{2}
A
(
t
)
=
[
s
(
t
)
]
2
\newline
(D)
[
A
(
t
)
]
2
=
[
s
(
t
)
]
2
+
[
s
(
t
)
]
2
[A(t)]^{2}=[s(t)]^{2}+[s(t)]^{2}
[
A
(
t
)
]
2
=
[
s
(
t
)
]
2
+
[
s
(
t
)
]
2
Get tutor help
The differentiable functions
x
x
x
and
y
y
y
are related by the following equation:
\newline
3
y
=
cos
(
x
)
3 y=\cos (x)
3
y
=
cos
(
x
)
\newline
Also,
d
y
d
t
=
5
\frac{d y}{d t}=5
d
t
d
y
=
5
.
\newline
Find
d
x
d
t
\frac{d x}{d t}
d
t
d
x
when
x
=
π
2
x=\frac{\pi}{2}
x
=
2
π
.
Get tutor help
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