Awall shaped like a trapezoid needs to be painted. the height of the wall is 16 feet and the bases are 20 feet and 24 feet. if one can of paint covers 20 square feet, how many whole cans of paint will the painter need to buy to paint the wall with one coat? the painter needs cans of paint.

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) = .

35 ! insert < , > , or = to make the sentence true. 12 __ 25 > =

An isosceles trapezoid with a perimeter of 42 inches. each of the congruent non parallel sides is 5 inches long, and the trapezoid is 3 inches tall. how long are the two parallel sides? a. 10 in, 22 in b. 16 in, 16 in c. 10 in, 16 in d. 12 in, 20 in