a= 6-2b-3c
then put that equation inside the first one and then the second one
you get -6*(6-2b-3c)+2b-3c=23
and 5*(6-2b-3c)+4b-3c=-12
14b-15c=59
14b-18c=18/*(-1)
--------------------
3b=41
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:
Since there isn't a line under the < sign, this means that we used a dotted or dashed line. The dotted or dashed line indicates that we do NOT include the boundary as part of the solution set.
Since y is isolated and we have a less than sign, this means we shade below the dashed/dotted boundary line. Specifically, the boundary line is the graph of y = 2x+1. This boundary line goes through (0,1) and (1,3). Again, points on this boundary line are NOT part of the solution set.
So in summary we have:
A dashed or dotted boundary line
The shaded region is below the dashed/dotted boundary line.
Answer:
50
Step-by-step explanation:
4,500÷90=50
and to check
90×50=4500