The solution set of a linear system whose augmented matrix is [a b c d] is the same as the solution set of Ax = d, where A = [a
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Answer:
Step-by-step explanation:
Let Y(s) be the Laplace transform Y=L{y(t)} of y(t)
Applying the Laplace transform to both sides of the differential equation and using the linearity of the transform, we get
L{y'' - 6y' + 9y} = L{0} = 0
(*) L{y''} - 6L{y'} + 9L{y} = 0 ; y(0)=4, y′(0)=2
Using the theorem of the Laplace transform for derivatives, we know that:
Replacing the initial values y(0)=4, y′(0)=2 we obtain
and our differential equation (*) gets transformed in the algebraic equation
Solving for Y(s) we get
Now, we brake down the rational expression of Y(s) into partial fractions
The numerator of the addition at the right must be equal to 4s-22, so
A(s - 3) + B = 4s - 22
As - 3A + B = 4s - 22
we deduct from here
A = 4 and -3A + B = -22, so
A = 4 and B = -22 + 12 = -10
It means that
and
By taking the inverse Laplace transform on both sides and using the linearity of the inverse:
we know that
and for the first translation property of the inverse Laplace transform
and the solution of our differential equation is
Answer:
The approximate circumference is 62.8 mm
Step-by-step explanation:
Circumference is basically the perimeter of the circle
Circumference is given by
Here the diameter is 20 mm, so the radius will be half of diameter = 10 mm
Substituting the given values, in above equation, we get -
Circumference =
mm
The approximate circumference is 62.8 mm