Strontium-90 90 Sr is a by-product of nuclear fission with a half-life of approximately 28.9 yr. After the Chernobyl nuclear reactor accident in 1986, large areas surrounding the site were contaminated with 90 Sr . if 10 μg (micrograms) of 90 Sr is present in a sample, the function A t = 10 1 2 t / 28.9 gives the amount A t in μg present after t years. Evaluate the function for the given values of t and interpret the meaning in context. Round to 3 decimal places if necessary. (See Example 5) a . A 28.9 b . A 57.8 c . A 100
Strontium-90 90 Sr is a by-product of nuclear fission with a half-life of approximately 28.9 yr. After the Chernobyl nuclear reactor accident in 1986, large areas surrounding the site were contaminated with 90 Sr . if 10 μg (micrograms) of 90 Sr is present in a sample, the function A t = 10 1 2 t / 28.9 gives the amount A t in μg present after t years. Evaluate the function for the given values of t and interpret the meaning in context. Round to 3 decimal places if necessary. (See Example 5) a . A 28.9 b . A 57.8 c . A 100
Solution Summary: The author calculates the remaining amounts of the radioactive element after different values of time t using the equation.
Strontium-90
90
Sr
is a by-product of nuclear fission with a half-life of approximately 28.9 yr. After the Chernobyl nuclear reactor accident in 1986, large areas surrounding the site were contaminated with
90
Sr
. if
10
μg
(micrograms) of
90
Sr
is present in a sample, the function
A
t
=
10
1
2
t
/
28.9
gives the amount
A
t
in μg
present after t years. Evaluate the function for the given values of t and interpret the meaning in context. Round to 3 decimal places if necessary. (See Example 5)
Write the given third order linear equation as an equivalent system of first order equations with initial values.
Use
Y1 = Y, Y2 = y', and y3 = y".
-
-
√ (3t¹ + 3 − t³)y" — y" + (3t² + 3)y' + (3t — 3t¹) y = 1 − 3t²
\y(3) = 1, y′(3) = −2, y″(3) = −3
(8) - (888) -
with initial values
Y
=
If you don't get this in 3 tries, you can get a hint.
Question 2
1 pts
Let A be the value of the triple integral
SSS.
(x³ y² z) dV where D is the region
D
bounded by the planes 3z + 5y = 15, 4z — 5y = 20, x = 0, x = 1, and z = 0.
Then the value of sin(3A) is
-0.003
0.496
-0.408
-0.420
0.384
-0.162
0.367
0.364
Question 1
Let A be the value of the triple integral SSS₂ (x + 22)
=
1 pts
dV where D is the
region in
0, y = 2, y = 2x, z = 0, and
the first octant bounded by the planes x
z = 1 + 2x + y. Then the value of cos(A/4) is
-0.411
0.709
0.067
-0.841
0.578
-0.913
-0.908
-0.120
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