If A is a 3 x 3 matrix and |A| = –2, then value of |A (adj A)| is (A)–2 (B) 2 (C) –8 (D) 8

c-   36

taxicab circle radius 9 has 4(9) = 36 points

substitute,

circle 1 = 5

circle 2 = 20

circle 3 = 44

in this case no matter how you add, you will always get a perfect square..

hence proved

happy to

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let's first notice, the center of the hyperbola is at the origin, since the focus points are equidistantly 3 units away from the origin, that also means that the distance "c" from the center to either focus is 3.

it should also be notable that, the traverse axis for this hyperbola is vertical, or namely over the y-axis and therefore, the positive fraction will be the one with y-variable.

the traverse "a" component or distance from the center to either vertex is not 2, is more than 2, so we can safely put it at √5 ≈ 2.24.

if i knew id you bro,, my bad