Answer:
The answer is a A. I just took a short test about it.
Step-by-step explanation:
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Answer with explanation:
The given differential equation is
y" -y'+y=2 sin 3x------(1)
Let, y'=z
y"=z'

Substituting the value of , y, y' and y" in equation (1)
z'-z+zx=2 sin 3 x
z'+z(x-1)=2 sin 3 x-----------(1)
This is a type of linear differential equation.
Integrating factor

Multiplying both sides of equation (1) by integrating factor and integrating we get


Answer:
x = 15°
y = 59°
Step-by-step explanation:
Angles at a right angle add up to 90°
41 + y = 90
y = 90 - 41
y = 59°
Angles in a triangle add up to 180°
41 + (2x - 9) + 7x + 13 = 180
41 + 13 - 9 + 2x + 7x = 180
45 + 9x = 180
9x = 180 - 45
9x = 135
x = 135/9
x = 15°