The perpendicular line to x-6y=2, and passing through (2, 4) is y=-6x+16
So you round to the second decimal place, giving you 43.59
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
Answer:
a. 1.2
Step-by-step explanation:
b. 0.4
