3teachers share 2 packs of paper equally how much does each teacher get

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) = .

This is a rational expression because the denominator contains a variable. this is a polynomial with 3 terms. this is a rational expression because the denominator contains a variable. this is a polynomial with 4 terms. this is a rational expression because the denominator contains a variable. this is a polynomial with 4 terms. this is a rational expression because the denominator contains a variable. this is a polynomial with 3 terms. this is a rational expression because the denominator contains a variable. this is a polynomial with 5 terms.

Solve the following: 12(x^2–x–1)+13(x^2–x–1)=25(x^2–x–1) 364x–64x=300x