Answer:
Perron–Frobenius theorem for irreducible matrices. Let A be an irreducible non-negative n × n matrix with period h and spectral radius ρ(A) = r. Then the following statements hold. The number r is a positive real number and it is an eigenvalue of the matrix A, called the Perron–Frobenius eigenvalue.
It is 40 40. 40 is the answer
(3a-4b)²
= 9a² - 12ab + 16b²
Answer:
see explanation
Step-by-step explanation:
h(x) + k(x) = x² + 1 + x - 2 = x² + x - 1
(h + k)(2) = 2² + 2 - 1 = 4 + 2 - 1 = 5
h(x) - k(x) = x² + 1 - (x - 2) = x² + 1 - x + 2 = x² -x + 3
(h - k)(3) = 3² - 3 + 3 = 9 - 3 + 3 = 9
h(2) = 2² + 1 = 4 + 1 = 5
k(3) = 3 - 2 = 1
3h(2) + 2k(3) = (3 × 5) + (2 × 1) = 15 + 2 = 17
5x6 = 30
3x7= 21
21+30 = 51
is this what you mean? if so, I'm glad to help