The answer would be A. When using Cramer's Rule to solve a system of equations, if the determinant of the coefficient matrix equals zero and neither numerator determinant is zero, then the system has infinite solutions. It would be hard finding this answer when we use the Cramer's Rule so instead we use the Gauss Elimination. Considering the equations:
x + y = 3 and <span>2x + 2y = 6
Determinant of the equations are </span>
<span>| 1 1 | </span>
<span>| 2 2 | = 0
</span>
the numerator determinants would be
<span>| 3 1 | . .| 1 3 | </span>
<span>| 6 2 | = | 2 6 | = 0.
Executing Gauss Elimination, any two numbers, whose sum is 3, would satisfy the given system. F</span>or instance (3, 0), <span>(2, 1) and (4, -1). Therefore, it would have infinitely many solutions. </span>
Answer:
y = 8 or y = 0
Step-by-step explanation:
Solve for y over the real numbers:
y^2 - 8 y = 0
Factor y from the left hand side:
y (y - 8) = 0
Split into two equations:
y - 8 = 0 or y = 0
Add 8 to both sides:
Answer: y = 8 or y = 0
Answer:
The answer is 27
Step-by-step explanation:
The reason being, you can solve this 2 ways.
1) Combine like terms, 7a-4a = 3a.
Now just substitute (a=9) into 3a.
3 * (9) = 27.
2) Or just substitute a = 9 into the original expression.
7 * (9) - 4 * (9)
= 63 - 36 = 27
(5 Neighbors) x (($1.00 Box) + (? Plant) + ($3.00 Gloves)) = 42.50
5 x (1 + x + 3) = 42.50
Combine in parens.
5 x (4 + x) = 42.50
Distribute.
20 + 5x = 42.50
Subtract 20 from both sides.
5x = 22.50
Divide both sides by 5.
x = 4.50
Thats the cost of each plant.
Check.
5 x (1 + 4.50 + 3)
5 x 8.50 = 42.50
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)
