Answer
(a) 
(b) 
Step-by-step explanation:
(a)
δ(t)
where δ(t) = unit impulse function
The Laplace transform of function f(t) is given as:

where a = ∞
=> 
where d(t) = δ(t)
=> 
Integrating, we have:
=> 
Inputting the boundary conditions t = a = ∞, t = 0:

(b) 
The Laplace transform of function f(t) is given as:



Integrating, we have:
![F(s) = [\frac{-e^{-(s + 1)t}} {s + 1} - \frac{4e^{-(s + 4)}}{s + 4} - \frac{(3(s + 1)t + 1)e^{-3(s + 1)t})}{9(s + 1)^2}] \left \{ {{a} \atop {0}} \right.](https://tex.z-dn.net/?f=F%28s%29%20%3D%20%5B%5Cfrac%7B-e%5E%7B-%28s%20%2B%201%29t%7D%7D%20%7Bs%20%2B%201%7D%20-%20%5Cfrac%7B4e%5E%7B-%28s%20%2B%204%29%7D%7D%7Bs%20%2B%204%7D%20-%20%5Cfrac%7B%283%28s%20%2B%201%29t%20%2B%201%29e%5E%7B-3%28s%20%2B%201%29t%7D%29%7D%7B9%28s%20%2B%201%29%5E2%7D%5D%20%5Cleft%20%5C%7B%20%7B%7Ba%7D%20%5Catop%20%7B0%7D%7D%20%5Cright.)
Inputting the boundary condition, t = a = ∞, t = 0:

It has been proven that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.
<h3>How to prove a Line Segment?</h3>
We know that in a triangle if one angle is 90 degrees, then the other angles have to be acute.
Let us take a line l and from point P as shown in the attached file, that is, not on line l, draw two line segments PN and PM. Let PN be perpendicular to line l and PM is drawn at some other angle.
In ΔPNM, ∠N = 90°
∠P + ∠N + ∠M = 180° (Angle sum property of a triangle)
∠P + ∠M = 90°
Clearly, ∠M is an acute angle.
Thus; ∠M < ∠N
PN < PM (The side opposite to the smaller angle is smaller)
Similarly, by drawing different line segments from P to l, it can be proved that PN is smaller in comparison to all of them. Therefore, it can be observed that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.
Read more about Line segment at; brainly.com/question/2437195
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Answer:
A, B, and D
Step-by-step explanation:
Answer:
x = 11, y = 22
Step-by-step explanation:
Using the tangent ratio in the right triangle and the exact value
tan60° =
, then
tan60° =
=
=
( multiply both sides by x )
11
= x ×
( divide both sides by
)
11 = x
Using Pythagoras' identity in the right triangle
y² = x² + (11
)² = 11² + 363 = 121 + 363 = 484 ( square root of both sides )
y =
= 22