Does anyone know the answers to these as well as the work shown for it?

The ksp expression for the system pbcl2(s) pb+2(aq) + 2 cl-(aq) is: [pbc12][pb+2][2cl-] [pb+2][cl-]2 [pb+2][2cl-]2 [pbcl2][pb+2][cl-]2 [pb+2][2cl-]2[pbcl2]

Long division the problem is 40 ➗ 348 = ?

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) = .