Answer:

Step-by-step explanation:

Answer:
The correct answer is C. 13
Step-by-step explanation:
It is 13 because all triangles equal 180
so 125+42+13=180
Answer:
x = 42
Step-by-step explanation:
The marked angles are supplementary, so their sum is 180°.
(2x +8) +(2x +4) = 180
4x +12 = 180 . . . . . . . . . simplify
x +3 = 45 . . . . . . . divide by 4 (because we can)
x = 42 . . . . . . subtract 3
_____
<em>Additional comment</em>
A "two-step" linear equation like this one is usually solved by subtracting the unwanted constant, then dividing by the coefficient of the variable. Here, we have done those steps in reverse order. This makes the numbers smaller and eliminates the coefficient of the variable. Sometimes I find it easier to solve the equation this way.
Take the homogeneous part and find the roots to the characteristic equation:

This means the characteristic solution is

.
Since the characteristic solution already contains both functions on the RHS of the ODE, you could try finding a solution via the method of undetermined coefficients of the form

. Finding the second derivative involves quite a few applications of the product rule, so I'll resort to a different method via variation of parameters.
With

and

, you're looking for a particular solution of the form

. The functions

satisfy


where

is the Wronskian determinant of the two characteristic solutions.

So you have




So you end up with a solution

but since

is already accounted for in the characteristic solution, the particular solution is then

so that the general solution is