Answer:
What are in the boxes?
Step-by-step explanation:
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:

Answer: The number pi is an irrational number
Step-by-step explanation:
Pi is irrational because it will never end and at the same time doesnt have a pattern in which the numbers repeat. Here are some of the numbers in pi 3.14159265359
You will receive a free beverage and free desert on your 60th visit.
Beverage- 1, 2, 3, 4, 5, 6
Visits- 12, 24, 36 ,48 ,60
Desert- 1, 2, 3 ,4
Visits- 15, 30, 45, 60
Notice how for beverages at 60 visits you get your 6th drink for free, and for deserts, at you 60 visits you receive your 4 free desert.
So, since both numbers have the same number (60) of visits which is their common multiple between the 2, you answer is 60.
Answer:
The area of rhombus PQRS is 120 m.
Step-by-step explanation:
Consider the rhombus PQRS.
All the sides of a rhombus are equal.
Hence, PQ = QR = RS = SP = 13 m
The diagonals PR and QS bisect each other.
Let the point at of intersection of the two diagonals be denoted by <em>X</em>.
Consider the triangle QXR.
QR = 13 m
XR = 12 m
The triangle QXR is a right angled triangle.
Using the Pythagorean theorem compute the length of QX as follows:
QR² = XR² + QX²
QX² = QR² - XR²
= 13² - 12²
= 25
QX = √25
= 5 m
The measure of the two diagonals are:
PR = 2 × XR = 2 × 12 = 24 m
QS = 2 × QX = 2 × 5 = 10 m
The area of a rhombus is:

Compute the area of rhombus PQRS as follows:


Thus, the area of rhombus PQRS is 120 m.