Question

The one linear pair and two supplementary angles with values of 1, 2, and 3 complete the proof. definition of supplementary angles congruent supplements theorem given transitive property of congruence linear pair theorem

Answer 1

The correct form of the partial fraction decomposition for (35-27x)/(4x²+28x) is (Ax+B)/(4x² + Cx) + (D/x + E/(x+7)). To find the values of A, B, C, D, and E, we need to rewrite the expression and equate the numerators and denominators. After solving the resulting system of equations, we can determine the values of the coefficients.

Step-by-step explanation:

The correct form of the partial fraction decomposition for (35-27x)/(4x²+28x) is (Ax+B)/(4x² + Cx) + (D/x + E/(x+7)).

To find the values of A, B, C, D, and E, we need to rewrite the expression and equate the numerators and denominators. After solving the resulting system of equations, we can determine the values of the coefficients.

1. First, factor the denominator: 4x² + 28x = 4x(x+7)
2. Write the expression as a sum of two fractions: (35-27x)/(4x²+28x) = A/(4x) + B/(x+7)
3. Combine the fractions and equate the numerators and denominators:(35-27x)/(4x²+28x) = (A(x+7) + B(4x))/(4x(x+7))
4. Equate the numerators: 35-27x = (A(x+7) + B(4x))
5. Equate the denominators: 4x²+28x = 4x(x+7)
6. Solve the resulting system of equations to find the values of A, B, C, D, and E.