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Math Problems
Precalculus
Velocity as a rate of change
E
=
43.66
ω
+
208.40
h
E=43.66 \omega+208.40 h
E
=
43.66
ω
+
208.40
h
\newline
The total energy,
E
E
E
, in joules of a metal cylinder with an angular velocity of
ω
\omega
ω
radians per second and a height of
h
h
h
meters above the ground is given by the equation. The height and angular velocity are independent. Which of the following expressions is the energy due to the angular velocity?
\newline
Choose
1
1
1
answer:
\newline
(A)
ω
\omega
ω
\newline
(B)
43
43
43
.
66
66
66
\newline
(C)
43.66
ω
43.66 \omega
43.66
ω
\newline
(D)
43.66
ω
+
208.40
43.66 \omega+208.40
43.66
ω
+
208.40
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A radioactive substance decays in such a way that the amount of mass remaining after
t
t
t
days is given by the function
m
(
t
)
=
12
e
−
0.017
t
m(t) = 12e^{-0.017t}
m
(
t
)
=
12
e
−
0.017
t
where
m
(
t
)
m(t)
m
(
t
)
is measured in kilograms. (a) Find the mass at time
t
=
0
t = 0
t
=
0
.
12
12
12
Correct: Your answer is correct. kg (b) How much of the mass remains after
50
50
50
days? (Round your answer to one decimal place.)
1.7
1.7
1.7
Incorrect: Your answer is incorrect. kg
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What are the critical points for the plane curve defined by the equations
\newline
x
(
t
)
=
cot
t
x(t)=\cot t
x
(
t
)
=
cot
t
,
y
(
t
)
=
sin
t
y(t)=\sin t
y
(
t
)
=
sin
t
, and
\newline
0
<
t
<
π
0 < t < \pi
0
<
t
<
π
? Write your answer as a list of values of
\newline
t
t
t
, separated by commas. For example, if you found
\newline
t
=
1
t=1
t
=
1
or
\newline
t
=
2
t=2
t
=
2
, you would enter
1
,
2
1,2
1
,
2
.
\newline
Provide your answer below:
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Q
=
A
−
I
L
Q=\frac{A-I}{L}
Q
=
L
A
−
I
\newline
The formula gives the quick assets ratio
Q
Q
Q
in terms of a company's current assets,
A
A
A
; inventories,
I
I
I
; and current liabilities,
L
L
L
. Which of the following equations correctly gives the inventories in terms of the quick assets ratio, the current assets, and the current liabilities?
\newline
Choose
1
1
1
answer:
\newline
(A)
I
=
Q
L
−
A
I=QL-A
I
=
Q
L
−
A
\newline
(B)
I
=
A
−
Q
L
I=A-QL
I
=
A
−
Q
L
\newline
(C)
I
=
L
(
Q
−
A
)
I=L(Q-A)
I
=
L
(
Q
−
A
)
\newline
(D)
I
=
L
(
A
−
Q
)
I=L(A-Q)
I
=
L
(
A
−
Q
)
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