Suppose you solved a second-order equation by rewriting it as a system and found two scalar solutions: y = e^5x and z = e^2x. Th
1 answer:
Answer:
The solutions are linearly independent because the Wronskian is not equal to 0 for all x.
The value of the Wronskian is
Step-by-step explanation:
We can calculate the Wronskian using the fundamental solutions that we are provided and their corresponding the derivatives, since the Wroskian is defined as the following determinant.
Thus replacing the functions of the exercise we get:
Working with the determinant we get
Thus we have found that the Wronskian is not 0, so the solutions are linearly independent.
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Work backwards.
s=ut+(at^2)/2
Subtract ut.
s-ut=(at^2)/2
Multiply by 2.
2(s-ut)=at^2
Divide by a.
2(s-ut)/a=t^2
Find the square root:
s=ut+(at^2)/2
Subtract ut.
s-ut=(at^2)/2
Multiply by 2.
2(s-ut)=at^2
Divide by a.
2(s-ut)/a=t^2
Find the square root:
Answer:
<h2>a) 0.38</h2><h2>b) 0.62</h2><h2>c) 0.78</h2><h2>d) 0.03</h2><h2>e) 0.02</h2><h2>f) 0.62</h2><h2>g) 0.38</h2>Step-by-step explanation:
a)
Probability of a owner to be moved = = 0.379 ≅ 0.38
b)
Probability of a renter to be moved = = 0.62
c)
Probability of a person who moved in the same state = = 0.779 ≅ 0.78
d)
Probability of a person who moved to a different country = = 0.033 ≅0.03
e)
Probability of a owner who moved to a different country = = 0.02
f)
Probability of a renter moving in the same state = = 0.615 ≅ 0.62
g)
Probability of an owner moved to different state = ≅0.38