We will have a system of equations, and we are going to solve it by using substitution.
Let's make the equations.
x+y+z=2100
z+y=x
Since y and z are equal, I think it is easier for us to have them as one variable for now. Let the value of z and y be a.
These are the equations now.
x+2a=2100
2a=x
Since the second equation is giving the value of x, let it replace the value of x also in the first equation.
2a+2a=2100
4a=2100
2100÷4= 525
so, x and y equal 525. Let's put this in our original equation.
x+525+525=2100
x+ 1,050= 2,100
x= 1,050
So, x=1,050, y= 525 =z.
5:6
The fraction 85/102 goes to 5/6 so it is 5:6 or 5/6...
We use P = i•e^rt for exponential population growth, where P = end population, i = initial population, r = rate, and t = time
P = 2•i = 2•15 = 30, so 30 = 15 [e^(r•1)],
or 30/15 = 2 = e^(r)
ln 2 = ln (e^r)
.693 = r•(ln e), ln e = 1, so r = .693
Now that we have our doubling rate of .693, we can use that r and our t as the 12th hour is t=11, because there are 11 more hours at the end of that first hour
So our initial population is again 15, and P = i•e^rt
P = 15•e^(.693×11) = 15•e^(7.624)
P = 15•2046.94 = 30,704
