(57-32)➗5.5 =x
The problem is just flipped.
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:
you literally just translate each shape by the number of squares specified.
Step-by-step explanation:
for example, move shape 1 5 squares to the right, then 4 down. it will be in the middle area of quadrant 1 (where it already is).
Answer:
9x^2 - 36x + 36 sq. ft. (the first option).
Step-by-step explanation:
Area of a square of side d = d^2.
So area of this square = (3x - 6)^2
= (3x - 6)(3x - 6)
= 3x(3x - 6) - 6(3x - 6)
= 9x^2 - 18x - 18x + 36
= 9x^2 - 36x + 36 sq ft.
1 gallon______8.354 lbs.
11.81 gallons______?
8.354 x 11.81 = 98.66
98.66 / 1 = 98.66 lbs