Plz i am really stuck on this question

What is the solution to the equation? k/6.4=8.7 2.3 5.568 15.1 55.68

With this question substitute t=3 and t=5 to determine if the two expressions are equivalent. 4(t + 3) 4t+12 which statements are true? check all that apply. a. the value of both expressions when t =5 is 32. b. the two expressions are not equivalent. c. the value of both expressions when t=3 is 15. d. the value of both expressions when t=5 is 23. e. the two expressions are equivalent. f. the value of both expressions when t=3 is 24.

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) = .