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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
=
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*}
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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
Taima was given this problem:
\newline
The radius
r
(
t
)
r(t)
r
(
t
)
of the base of a cone is decreasing at a rate of
2
2
2
centimeters per minute. The height of the cone is fixed at
12
12
12
centimeters. At a certain instant
t
0
t_{0}
t
0
, the radius is
13
13
13
centimeters. What is the rate of change of the surface area
S
(
t
)
S(t)
S
(
t
)
of the cone at that instant?
\newline
Which equation should Taima use to solve the problem?
\newline
Choose
1
1
1
answer:
\newline
(A)
S
(
t
)
=
π
[
r
(
t
)
]
2
+
2
π
⋅
r
(
t
)
⋅
12
S(t)=\pi[r(t)]^{2}+2 \pi \cdot r(t) \cdot 12
S
(
t
)
=
π
[
r
(
t
)
]
2
+
2
π
⋅
r
(
t
)
⋅
12
\newline
(B)
S
(
t
)
=
π
[
r
(
t
)
]
2
⋅
12
S(t)=\pi[r(t)]^{2} \cdot 12
S
(
t
)
=
π
[
r
(
t
)
]
2
⋅
12
\newline
(C)
S
(
t
)
=
π
[
r
(
t
)
]
2
+
π
⋅
r
(
t
)
[
r
(
t
)
]
2
+
1
2
2
S(t)=\pi[r(t)]^{2}+\pi \cdot r(t) \sqrt{[r(t)]^{2}+12^{2}}
S
(
t
)
=
π
[
r
(
t
)
]
2
+
π
⋅
r
(
t
)
[
r
(
t
)
]
2
+
1
2
2
\newline
(D)
S
(
t
)
=
π
[
r
(
t
)
]
2
⋅
12
3
S(t)=\frac{\pi[r(t)]^{2} \cdot 12}{3}
S
(
t
)
=
3
π
[
r
(
t
)
]
2
⋅
12
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The function
b
(
t
)
b(t)
b
(
t
)
gives the number of books sold by a store by time
t
t
t
(in days) of a given year.
\newline
What does
∫
45
50
b
′
(
t
)
d
t
\int_{45}^{50} b^{\prime}(t) d t
∫
45
50
b
′
(
t
)
d
t
represent?
\newline
Choose
1
1
1
answer:
\newline
(A) The number of books sold between day
45
45
45
and day
50
50
50
\newline
(B) The number of days it takes to sell
50
50
50
books
\newline
(C) The change in the rate of selling books between
t
=
45
t=45
t
=
45
and
t
=
50
t=50
t
=
50
\newline
(D) The total number of books sold by day
50
50
50
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