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Mathematics, 21.06.2019 19:50
Prove (a) cosh2(x) β sinh2(x) = 1 and (b) 1 β tanh 2(x) = sech 2(x). solution (a) cosh2(x) β sinh2(x) = ex + eβx 2 2 β 2 = e2x + 2 + eβ2x 4 β = 4 = . (b) we start with the identity proved in part (a): cosh2(x) β sinh2(x) = 1. if we divide both sides by cosh2(x), we get 1 β sinh2(x) cosh2(x) = 1 or 1 β tanh 2(x) = .
Answers: 3
Mathematics, 21.06.2019 22:10
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.
Answers: 2
Mathematics, 21.06.2019 23:30
Solve the following: 12(x^2βxβ1)+13(x^2βxβ1)=25(x^2βxβ1) 364xβ64x=300x
Answers: 1
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