Which is a stretch of an exponential growth function? f(x) = Two-thirds (two-thirds) Superscript x f(x) = Three-halves (two-thir
2 answers:
Answer:
f(x) = Three-halves (three-halves) Superscript x
f(x) = Two-thirds (three-halves) Superscript x
Step-by-step explanation:
Since, a function in the form of
Where, a and b are any constant,
is called exponential function,
There are two types of exponential function,
- Growth function : If b > 1,
- Decay function : if 0 < b < 1,
Since, In
Thus, it is a decay function.
in
Thus, it is a decay function.
in
Thus, it is a growth function.
in
Thus, it is a growth function.
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Answer:
the P matrix you are looking for is P=(1/) · [[1 1 0 0],[1 -1 0 0],[0 0 1 1],[0 0 1 -1]]
Step-by-step explanation:
Answer:
For an orthogonal diagonalization of any matrix you have to:
1º) Find the matrix eigenvalues in a set order.
2º) Find the eigenvectors of each respective eigenvalues.
Tip: You can write the matrix A like A = D P
3º) D is the diagonal matrix with each eigenvalue (in order) in the diagonal.
4º) Write P as the normalized eigenvectors in order (in columns).
Tip 2: Remember, ·P = I, so if A =
D P, then:
P A = P
D P
= I D I = D
So the P we are looking for is the of the diagonalization.
Tip 3: In this case, A is a block matrix with null nondiagonal submatrixes, therefore its eigenvalues can be calculated by using the diagonal submatrixes. The problem is reduced to calculate the eigenvalues of A₁₁ = A ₂₂ = [[5 3],[3 5]]
Solving:
1º)the eigenvalues of A₁₁ are {8,2}, therefore the D matrix is
2º) the eigenvectors of A₁₁ are P₈= {[1 1]T} P₂= {[1 -1]T}, therefore normalizing the eigenvectors you obtain P = (1/) · [[1 1 0 0],[1 -1 0 0],[0 0 1 1],[0 0 1 -1]] (you can see that P =
in this case).
As said in "tip 2": P A = P
D P
= I D I = D
So the P obtained is the one you are looking for.
