Find three consecutive odd integers whose sum is 117

First odd integer: 2k+1 (note: using 2k+1 insures it is an odd integer)

Second odd integer: 2k+3

Third odd integer: 2k+5

First + Second + Third = 117

2k+1 + 2k+3 + 2k+5 = 117 substituted for First, Second, and Third

6k + 9 = 117 added like terms

6k = 108 subtracted 7 from both sides

k = 18

First: 2k + 1 = 2(18) + 1 = 36 + 1 = 37

Second: 2k + 3 = 2(18) + 3 = 36 + 3 = 39

Third: 2k + 5 = 2(18) + 5 = 36 + 5 = 41

37, 39, 41

256π units²   (answer b)

step-by-step explanation:

the formula for the surface area of a sphere of radius r

is             a = 4πr²

and so, if r = 8 units, then the surface area is:

a = 4π(8 units)² = 256π units²   (answer b)

step-by-step explanation: