If two similar polygons have a scale factor of 1, what can you conclude? A. They are sometimes congruent.
B. They are always congruent.
C. They are never congruent.
D. They would only be congruent if they were regular polygons.

Prove (a) cosh2(x) − sinh2(x) = 1 and (b) 1 − tanh 2(x) = sech 2(x). solution (a) cosh2(x) − sinh2(x) = ex + e−x 2 2 − 2 = e2x + 2 + e−2x 4 − = 4 = . (b) we start with the identity proved in part (a): cosh2(x) − sinh2(x) = 1. if we divide both sides by cosh2(x), we get 1 − sinh2(x) cosh2(x) = 1 or 1 − tanh 2(x) = .

Can someone me and explain..will award brainlest!

2pointswhat is the best next step in the construction of the perpendicular bisector ofab?