(12.5*2*Pi) / 8 = 3.125*Pi [1/s]
Answer:
http://fumacrom.com/8rEB
Step-by-step explanation:
http://fumacrom.com/8rEB
Answer:
Answers In Images Below \/
Step-by-step explanation:
At the end of the zeroth year, the population is 200.
At the end of the first year, the population is 200(0.96)¹
At the end of the second year, the population is 200(0.96)²
We can generalise this to become at the end of the nth year as 200(0.96)ⁿ
Now, we need to know when the population will be less than 170.
So, 170 ≤ 200(0.96)ⁿ
170/200 ≤ 0.96ⁿ
17/20 ≤ 0.96ⁿ
Let 17/20 = 0.96ⁿ, first.
log_0.96(17/2) = n
n = ln(17/20)/ln(0.96)
n will be the 4th year, as after the third year, the population reaches ≈176
21 times x to the fifth (where x is the ratio between values) equals 1240029
1240029/21=59049
59049^(1/5)=9
so the values are 21, 21*9, 21*9^2, 21*9^3, 21*9^4, 1240029
the sum of these values is 1395030