Please who dont know the ans ignore it ... please quik help​

You drop a rubber ball off the roof of a 50 meter high building onto a paved parking lot. it bounces back up with every bounce, but not quite all the way back up to you. after the first bounce it bounces back only 80 percent of the distance it was dropped from. the pattern continues, meaning after every bounce it comes up to just 80 percent of the previous maximum height. so if before the first bounce the height is 50 meters, what height does the ball reach after the fifth bounce? round your answer to one decimal place and chose the correct response from the choices below:

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) = .

Given: klmn is a parallelogram m? n=3m? k, lf ? kn , ld ? nm kf=2 cm, fn=4 cm find: lf, ld