Answer:
Suppose a population of rodents satisfies the differential equation dP 2 kP dt = . Initially there are P (0 2 ) = rodents, and their number is increasing at the rate of 1 dP dt = rodent per month when there are P = 10 rodents.
How long will it take for this population to grow to a hundred rodents? To a thousand rodents?
Step-by-step explanation:
Use the initial condition when dp/dt = 1, p = 10 to get k;
Seperate the differential equation and solve for the constant C.
You have 100 rodents when:
You have 1000 rodents when:
Answer:
The length from one point to another that is a portion of the curve of the circle is called <u>arc</u>
Answer:
See attachment
Step-by-step explanation:
First solve this like a normal equation:
-3x - 6 > 9
Add 6 to both sides:
-3x > 9 + 6
-3x > 15
Divide by -3 and remember to switch the inequality sign:
x < 15/(-3)
x < -5
Think about the graph of x = -5. It's a vertical line crossing the y-axis at x = -5. Now, we have x is less than -5. That means all the values less than -5 should be viable solutions. So, shade the part of the graph to the left of the vertical line.
Also, since we have x < -5 and not x ≤ -5, the line should be dotted.
See graph attached.
<em>~ an aesthetics lover</em>
Answer:
(5,-3) in the 4th quadrant
Step-by-step explanation:
Amount borrowed, P = 13000
Interest, i = 0.07/12 per month
number of periods (months), n = 4*12 = 48
Monthly payment,
=
311.30 (to the nearest cent)
