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Algebra 2
Solve quadratic equations: word problems
The atmospheric pressure of the air changes with height above sea level. The rate of change of the height above sea level for a given air pressure can be measured by the differentiable function
f
(
p
)
f(p)
f
(
p
)
, in kilometers per psi, where
p
p
p
is measured in psi. What are the units of
∫
2
5
f
(
p
)
d
p
?
\int_{2}^{5} f(p) d p ?
∫
2
5
f
(
p
)
d
p
?
\newline
kilometers
\newline
psi
\newline
kilometers / psi
\newline
psi / kilometer
\newline
kilometers
/
p
s
i
2
/ \mathrm{psi}^{2}
/
psi
2
\newline
psi / kilometer
2
{ }^{2}
2
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The atmospheric pressure of the air changes with height above sea level. The rate of change of the air pressure at a given height above sea level can be measured by the differentiable function
f
(
h
)
f(h)
f
(
h
)
, in psi per foot, where
h
h
h
is measured in feet. What are the units of
f
′
(
h
)
f^{\prime}(h)
f
′
(
h
)
?
\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
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The temperature of a turkey in an oven is measured by the differentiable function
f
f
f
, where
f
(
t
)
f(t)
f
(
t
)
is measured in degrees Fahrenheit and
t
t
t
is measured in minutes. What are the units of
f
′
′
(
t
)
f^{\prime \prime}(t)
f
′′
(
t
)
?
\newline
degrees Fahrenheit
\newline
degrees Fahrenheit / minute
\newline
degrees Fahrenheit / minute
2
{ }^{2}
2
\newline
minutes
\newline
minutes / degree Fahrenheit
\newline
minutes / degree Fahrenheit
2
{ }^{2}
2
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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 minutes. What are the units of
∫
3
6
f
′
(
t
)
d
t
\int_{3}^{6} f^{\prime}(t) d t
∫
3
6
f
′
(
t
)
d
t
?
\newline
liters
\newline
minutes
\newline
liters / minute
\newline
minutes / liter
\newline
liters / minute
2
{ }^{2}
2
\newline
minutes / liter
2
{ }^{2}
2
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Ori was given this problem:
\newline
The radius
r
(
t
)
r(t)
r
(
t
)
of the base of a cone is increasing at a rate of
10
10
10
meters per second. The height
h
(
t
)
h(t)
h
(
t
)
of the cone is fixed at
6
6
6
meters. At a certain instant
t
0
t_{0}
t
0
, the radius is
1
1
1
meter. What is the rate of change of the volume
V
(
t
)
V(t)
V
(
t
)
of the cone at that instant?
\newline
Which equation should Ori use to solve the problem?
\newline
Choose
1
1
1
answer:
\newline
(A)
V
(
t
)
=
π
[
r
(
t
)
]
2
+
2
π
⋅
r
(
t
)
⋅
h
(
t
)
V(t)=\pi[r(t)]^{2}+2 \pi \cdot r(t) \cdot h(t)
V
(
t
)
=
π
[
r
(
t
)
]
2
+
2
π
⋅
r
(
t
)
⋅
h
(
t
)
\newline
(B)
V
(
t
)
=
π
[
r
(
t
)
]
2
⋅
h
(
t
)
V(t)=\pi[r(t)]^{2} \cdot h(t)
V
(
t
)
=
π
[
r
(
t
)
]
2
⋅
h
(
t
)
\newline
(C)
V
(
t
)
=
π
[
r
(
t
)
]
2
+
π
⋅
r
(
t
)
[
r
(
t
)
]
2
+
[
h
(
t
)
]
2
V(t)=\pi[r(t)]^{2}+\pi \cdot r(t) \sqrt{[r(t)]^{2}+[h(t)]^{2}}
V
(
t
)
=
π
[
r
(
t
)
]
2
+
π
⋅
r
(
t
)
[
r
(
t
)
]
2
+
[
h
(
t
)
]
2
\newline
(D)
V
(
t
)
=
π
[
r
(
t
)
]
2
⋅
h
(
t
)
3
V(t)=\frac{\pi[r(t)]^{2} \cdot h(t)}{3}
V
(
t
)
=
3
π
[
r
(
t
)
]
2
⋅
h
(
t
)
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The cumulative profit a business has earned is changing at a rate of
r
(
t
)
r(t)
r
(
t
)
dollars per day (where
t
t
t
is the time in days). In the first
30
30
30
days, the business earned a cumulative profit of
$
1700
\$ 1700
$1700
.
\newline
What does
1700
+
∫
30
90
r
(
t
)
d
t
1700+\int_{30}^{90} r(t) d t
1700
+
∫
30
90
r
(
t
)
d
t
represent?
\newline
Choose
1
1
1
answer:
\newline
(A) The cumulative profit the business has earned as of day
90
90
90
\newline
(B) The change in the cumulative profit between days
30
30
30
and
90
90
90
\newline
(C) The rate at which the cumulative profit was increasing when
t
=
90
t=90
t
=
90
.
\newline
(D) The time it takes for the cumulative profit to increase another
$
1700
\$ 1700
$1700
after the first
30
30
30
days
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