I have no clue what the answer is
Answer: Although A and AT have the same eigenvalues, they may actually have different eigenvectors for that particular eigenvalue. Therefore A and AT have at least one eigenvector that aren't in common.
Step-by-step explanation:
These two attachment explains the provings and counterexamples briefly and simply
5 cups.
To go from 1/2 to 2 cups of raisins you multiply by 4.
So 1 1/4 is really 5/4. That times 4 is 20/4 which is 5.
u(x) = -2x², v(x)= 1/x
(u o v)(x) = u(v(x)) = -2(1/x)²= -2/x²
We can see that domain for x is going to bee all real numbers except 0. From the equation above we can see that graph of the function (u o v)(x) = -2/x² has horizontal asymptote y=0, because degree of numerator is less than degree of denominator ( (u o v)(x) = -2x⁰/x² ).
x² is always going to be positive, so range is going to be all negative numbers.
y<0, or y∈(0,-∞)
I think that the base of this parallelogram is 13.6 cm. I may be wrong though. Double check the work. Divide area by width and you'll get the length.