HELP PLEASE?? Identify the 19th term of a geometric sequence where a1 = 14 and a9 = 358.80. Round the common ratio and 19th term to the nearest hundredth.

Question

anonymous · Answer

r^8 = (358.8 / 14)
r = (358.8 / 14)^(1/8)

14 = m * (358.8 / 14)^(1/8)
14 / (358.8 / 14)^(1/8) = m

(14 / (358.8 / 14)^(1/8)) * ((358.8 / 14)^(1/8))^19 =>
14 * ((358.8 / 14)^(1/8))^18 =>
14 * (358.8 / 14)^(18/8) =>
14 * (358.8 / 14)^2 * (358.8 / 14)^(1/4) =>
20689.878139502656162447672308237

anonymous · Answer

Sorry can you help me with on emore?

anonymous · Answer

A bacteria culture begins with 15 bacteria which double in amount at the end of every hour. How many bacteria exist in the culture at the end of the 12th hour?

anonymous · Answer

After 1 hour you have 15(2) = 30
After 2 hours you have 15(2)(2) = 60
After 3 hours you have 15(2)^3 = 120

and so on.

So after 12 hours you have 15(2)^12 = 15(4096) = 61440