Prove that the conjugate transpose of a matrix product is the product of the conjugate transposes in the opposite order, i. e., (ab)* = b'a . do this in general using indices, i. e., show that ((ab)*); k = (b*a") is using the definitions of conjugate transpose and matrix multiplication and properties of the conjugate.

the anwser is d) 75°

step-by-step explanation:

the answer would be 44.79

plz mark brainliest

step-by-step explanation:

step-by-step explanation:

given that ab is divided into equal parts as

ac, cd, de, ef, fg, gh, hi, ib.

from this we find that f is the midpoint of ab.

so using mid point formula we find

f = (3,7)

similarly considering fb, we get

h = mid point of fb = ((6,6) and g, mid point of fh is

g = (9/2, 13/2)

i is mid point of hb

so i =(15/2,11/2)

answer: 7. if the functions had a negative leading coefficient and is of even degree, which statement about the graph is true?

step-by-step explanation: