The linear function is represented by the equation y = Negative three-fifthsx – 2. What can be determined of this equation written in slope-intercept form? Check all that apply.

Simplify. −4x^2 (5x^4−3x^2+x−2) −20x^6−12x^4+8x^3−8x^2 −20x^6+12x^4−4x^3+8x^2 −20x^8+12x^4−4x^2+8x −20x^6+12x^4+4x^3−8x^2

The length of a shadow of the building is 36m. the length of the pole, which is perpendicular to the ground, is 1.9m. the length of the pole's shadow is 1.52m. what is the height of the building?

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) = .