Answer:
y = 1.5x + 3.2009
Step-by-step explanation:
I plotted your points and drew a linear regression line.
Its equation is
y = 1.5x + 3.2009
The regression line ignores the fact that the points fall along a smooth curve.
A parabolic regression line fits the data much better.
The equation for that line is
y = - 0.1647x² + 4.1357x - 5.6946
It is the equation for a vertical parabola opening downward.
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:
Answer:
48 in
Step-by-step explanation:
The diagonal of the inscribed square equals the diameter of the circle.
1. Calculate the diagonal of the square
d = 2r = 2 × 6√2 in =12√2 in
2. Calculate the side of the square
∆ACB is an isosceles right triangle.
We can use Pythagoras' Theorem to calculate the length of a side s.
3. Calculate the perimeter of the square
P = 4s = 4 × 12 in = 48 in
The perimeter of the square is 48 in.
