Answer:
Check the explanation
Step-by-step explanation:
where the letter D is the diagonal matrix with diagonal entries λ1,…,λn. Now let's assume V is invertible, that is, this particular given eigenvectors are linearly independent, you get M=VDV−1.
Kindly check the attached image below to see the step by step explanation to the question above.
You can either scale down so you can get to 48 hours or just multiply from 20 to 48 and I'll explain both ways.
Scaling down:
You can find how many hours it takes to get to 1 gallon in the bucket by dividing
20÷5=4; It takes 4 hours to fill it up to one gallon.
You can now divide 48 by 4 to see how many gallons it'll take up to 48 hours.
48÷4=12
Quicker:
Divide 48 by 20;
48÷20=2.4
Now multiply since it filled it up to 5 gallons.
2.4×5=12
It'll fill 12 gallons in 48 hours.
Tell me if this helps!!
Data:
x: number of months
y: tree's height
Tipical grow: 0.22
Fifteen months into the observation, the tree was 20.5 feet tall: x=15 y=20.5ft (15,20.5)
In this case the slope (m) or rate of change is the tipical grow.
m=0.22
To find the line's slope-intercep equation you use the slope (m) and the given values of x and y (15 , 20.5) in the next formula to find the y-intercept (b):

Use the slope(m) and y-intercept (b) to write the equation:

A) This line's slope-intercept equation is: y=0.22x+17.2
B) To find the height of the tree after 29 months you substitute in the equation the x for 29 and evaluate to find the y:

Then, after 29 months the tree would be 23.58 feet in height
C) In this case as you have the height and need to find the number of moths you substitute the y for 29.96feet and solve the equation for x, as follow:

Then, after 58 months the tree would be 29.96feet tall
Find the median of each set of data. 12, 8, 6, 4, 10, 1 7 6, 3, 5, 11, 2, 9, 5, 0 5 30, 16, 49, 25
professor190 [17]
Answer:
first 7 then 5 and 27.5
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)