Log5 x– log5 25 = 7
A. 59 = x
=
B. 55=x
C. 75 = x
D. 79=x

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) = .

|y + 2| > 6 {y|y < -8 or y > 4} {y|y < -6 or y > 6} {y|y < -4 or y > 4}

A3cm×3cm\greend{3\,\text{cm} \times 3\,\text{cm}} 3cm×3cm start color greend, 3, space, c, m, times, 3, space, c, m, end color greend rectangle sits inside a circle with radius of 4cm\blued{4\,\text{cm}} 4cm start color blued, 4, space, c, m, end color blued . what is the area of the shaded region? round your final answer to the nearest hundredth.