Answer:
Step-by-step explanation:
The growth of the population can be modeled by the following differential equation:
Where r is the growth rate, P is the population, and t is the time measures in months.
I am going to solve the above differential equation with the separation of variables method.
Integrating both sides:
Where P(0) is the initial condition
We need to isolate P in this equation, so we do this
So
The problem states that P(0)=3000, so:
The problem wants us to find the value of r:
It states that the population doubles in size from 3000 to 6000 in a 6- month period, meaning that P(6) = 6000. So
To isolate 6r, we apply ln to both sides.
r = 0.1155
The particular solution to the differential equation with the initial condition P(0)=3000 is:
A translation moves the graph
A rotation spins it
A reflection flips it
Answer:
So if 14% are over 65, and 320 residents are over 65, then we can say that 14% of the town is 320. Then, we can divide by 14 to find 1%, which is 22.857142857 Now we can multiply this by 100 to get 2285.7142857.
Step-by-step explanation:
Answer:
0 < x ≤ 12 and 0 < y ≤ 36
Step-by-step explanation:
Here, x represents the number of female gazelles and y represents the number of male gazelles.
The zoo only has room for 12 female gazelles.
∵ The number of rooms must be more than or equal to the total female gazelles ,
12 ≥ x
Also, number of animals can not be negative,
And, it must be greater than 0.
⇒ 0 < x ≤ 12,
⇒ 3(0) < 3x ≤ 3(12)
⇒ 0 < 3x ≤ 36
∵ Number of males gazelles = 3 × number of female gazelles
⇒ y = 3x
⇒ 0 < y ≤ 36
Hence, the constraints to represent a thriving population of gazelles at the zoo are,
0 < x ≤ 12,
0 < y ≤ 36
Answer:
-12.5%
Step-by-step explanation:
The percentage change in your time can be computed using ...
pct change = ((new value)/(old value) -1) × 100%
= (28/32 -1) × 100%
= (0.875 -1) × 100% = -12.5%
The time to finish level 2 decreased by 12.5%.
