If the sides of a cube are 4 inches long, then its volume is a) 64 in b) 64 in2 c) 64 in3 d) 64 in4

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) = .

Ihave a triangle with a 60 degree angle. lets use d for that angle. what are the values of e and f if d is 60 degrees? and what do you notice about the values of e and f?

Each of the following data sets has a mean of x = 10. (i) 8 9 10 11 12 (ii) 7 9 10 11 13 (iii) 7 8 10 12 13 (a) without doing any computations, order the data sets according to increasing value of standard deviations. (i), (iii), (ii) (ii), (i), (iii) (iii), (i), (ii) (iii), (ii), (i) (i), (ii), (iii) (ii), (iii), (i) (b) why do you expect the difference in standard deviations between data sets (i) and (ii) to be greater than the difference in standard deviations between data sets (ii) and (iii)? hint: consider how much the data in the respective sets differ from the mean. the data change between data sets (i) and (ii) increased the squared difference îł(x - x)2 by more than data sets (ii) and (iii). the data change between data sets (ii) and (iii) increased the squared difference îł(x - x)2 by more than data sets (i) and (ii). the data change between data sets (i) and (ii) decreased the squared difference îł(x - x)2 by more than data sets (ii) and (iii). none of the above