Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 11 - Test - Page 669: 15

Answer

Fourth swing distance: $33.75$ cm Total of first four swing distances: $218.75$ cm

Work Step by Step

Swing is 80 cm on first swing, successive swings are 3/4 of previous swing Looking for fourth swing distance and sum of first four swing distances $S_3 + $ 4th swing distance $=S_4$ $S_4 = (a_1)*(1-r^n)/(1-r)$ $a_1 = 80$ $r = .75$ $n=4$ $S_4 = (a_1)*(1-r^n)/(1-r)$ $S_4 = 80 *(1-.75^4)/(1-.75)$ $S_4 = 80*(1-.31640625)/.25$ $S_4 = 80*.68359375/.25$ $S_4= 80*2.734375$ $S_4 = 218.75$ $a_1 = 80$ $r = .75$ $n=3$ $S_3 = (a_1)*(1-r^n)/(1-r)$ $S_3 = 80*(1-.75^3)/(1-.75)$ $S_3 = 80 *(1-.421875)/.25$ $S_3=80*.578125/.25$ $S_3 = 80*2.3125$ $S_3 = 185$ $S_4 – S_3 =$ distance of fourth swing $218.75-185$ $33.75$ cm is distance of fourth swing
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