These angles should add to 180 degrees because the angles are on a straight line so you should use the formula x=180-90-38. Therefore, x should be 52 degrees
Answer:
Step-by-step explanation:
Let's use the definition of the Laplace transform and the identity given:![\mathcal{L}[t \cos 5t]=(-1)F'(s)](https://tex.z-dn.net/?f=%5Cmathcal%7BL%7D%5Bt%20%5Ccos%205t%5D%3D%28-1%29F%27%28s%29)
with ![F(s)=\mathcal{L}[\cos 5t]](https://tex.z-dn.net/?f=F%28s%29%3D%5Cmathcal%7BL%7D%5B%5Ccos%205t%5D)
.
Now, 
. Using integration by parts with u=e^(-st) and dv=cos(5t), we obtain that 
.
Using integration by parts again with u=e^(-st) and dv=sin(5t), we obtain that
.
Solving for F(s) on the last equation, 
, then the Laplace transform we were searching is 
8x - 7 = 3x + 9
8x - 3x = 9 + 7
5x = 16
x = 16/5 = 3 1/5
Ok, first group x terms
f(x)=(x²+4x)-8
factor out quadratic coefient (no need but that's the step)
f(x)=1(x²+4x)-8
take 1/2 of the linear coefient and square it
4/2=2, (2)²=4
add positive and negative of it insides the parenthasees
f(x)=1(x²+4x+4-4)-8
factor perfect square
f(x)=1((x+2)²-4)-8
distribute
f(x)=1(x+2)²-4-8
f(x)=1(x+2)²-12
and, now if we wanted to find the x intercepts where f(x)=0 then
0=1(x+2)²-12
12=(x+2)²
+/-2√3=x+2
-2+/-2√3=x
x=-2+2√3 or -2-2√3
that is where the x intercept are
and completed square form is
f(x)=(x+2)²-12
