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:
B) 7 feet is the median (center) and 5 is the range (spread)
Step-by-step explanation:
spread is the difference between the largest and smallest numbers in the data set (9 - 4 = 5 in this case)
zero
double - is actually plus
<u>Work 1:</u>
Successful Percent:
15 divided by 20 equals .75
.75 times 100 is equal to 75
75%
Not successful Percent:
100% minus 75% equals 25%
<em>25%</em>
<u>Work 2:</u><u />
20 minus 15 equals 5
5 divided by 20 is equal to .25
.25 times 100 is equal to 25
<em>25%</em>
Answer:
Step-by-step explanation:
Here's how you convert:
The little number outside the radical, called the index, serves as the denominator in the rational power, and the power on the x inside the radical serves as the numerator in the rational power on the x.
A couple of examples:
![\sqrt[3]{x^4}=x^{\frac{4}{3}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%5E4%7D%3Dx%5E%7B%5Cfrac%7B4%7D%7B3%7D)
![\sqrt[5]{x^7}=x^{\frac{7}{5}](https://tex.z-dn.net/?f=%5Csqrt%5B5%5D%7Bx%5E7%7D%3Dx%5E%7B%5Cfrac%7B7%7D%7B5%7D)
It's that simple. For your problem in particular:
is the exact same thing as ![\sqrt[3]{7^1}=7^{\frac{1}{3}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B7%5E1%7D%3D7%5E%7B%5Cfrac%7B1%7D%7B3%7D)