Pretty difficult problem, but that’s why I’m here.
First we need to identify what we’re looking for, which is t. So now plug 450k into equation and solve for t.
450000 = 250000e^0.013t
Now to solve this, we need to remember this rule: if you take natural log of e you can remove x from exponent. And natural log of e is 1.
Basically ln(e^x) = xln(e) = 1*x
So knowing this first we need to isolate e
450000/250000 = e^0.013t
1.8 = e^0.013t
Now take natural log of both
Ln(1.8) = ln(e^0.013t)
Ln(1.8) = 0.013t*ln(e)
Ln(1.8) = 0.013t * 1
Now solve for t
Ln(1.8)/0.013 = t
T= 45.21435 years
Now just to check our work plug that into original equation which we get:
449999.94 which is basically 500k (just with an error caused by lack of decimals)
So our final solution will be in the 45th year after about 2 and a half months it will reach 450k people.
If f(x) is an inverse of g(x),
when
f(x)=y
g(y)=x
aka
f(g(x))=x
g(f(x))=x
basically, the values should be swiched
example
f(x)=
(1,2)
(2,3)
(4,5)
then g(x)=
(2,1)
(3,2)
(5,4)
Answer:
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Step-by-step explanation:
Answer:
Step-by-step explanation:
(A) The difference between an ordinary differential equation and an initial value problem is that an initial value problem is a differential equation which has condition(s) for optimization, such as a given value of the function at some point in the domain.
(B) The difference between a particular solution and a general solution to an equation is that a particular solution is any specific figure that can satisfy the equation while a general solution is a statement that comprises all particular solutions of the equation.
(C) Example of a second order linear ODE:
M(t)Y"(t) + N(t)Y'(t) + O(t)Y(t) = K(t)
The equation will be homogeneous if K(t)=0 and heterogeneous if 
Example of a second order nonlinear ODE:
(D) Example of a nonlinear fourth order ODE:
![K^4(x) - \beta f [x, k(x)] = 0](https://tex.z-dn.net/?f=K%5E4%28x%29%20-%20%5Cbeta%20f%20%5Bx%2C%20k%28x%29%5D%20%3D%200)
Answer:
a
Step-by-step explanation:
