Answer: below in picture
explanation: also in picture
![f(t)=\begin{cases}0&\text{for }t](https://tex.z-dn.net/?f=f%28t%29%3D%5Cbegin%7Bcases%7D0%26%5Ctext%7Bfor%20%7Dt%3C0%5C%5C-5%26%5Ctext%7Bfor%20%7D0%5Cle%20t%3C1%5C%5C-6%26%5Ctext%7Bfor%20%7D1%5Cle%20t%3C7%5C%5C6%26%5Ctext%7Bfor%20%7Dt%5Cge7%5Cend%7Bcases%7D)
Recall that
![u(t)=\begin{cases}0&\text{for }t](https://tex.z-dn.net/?f=u%28t%29%3D%5Cbegin%7Bcases%7D0%26%5Ctext%7Bfor%20%7Dt%3C0%5C%5C1%26%5Ctext%7Bfor%20%7Dt%5Cge0%5Cend%7Bcases%7D)
Take it one piece at a time. For
![t\ge0](https://tex.z-dn.net/?f=t%5Cge0)
, we can scale
![u(t)](https://tex.z-dn.net/?f=u%28t%29)
by -5:
![-5u(t)=\begin{cases}0&\text{for }t](https://tex.z-dn.net/?f=-5u%28t%29%3D%5Cbegin%7Bcases%7D0%26%5Ctext%7Bfor%20%7Dt%3C0%5C%5C-5%26%5Ctext%7Bfor%20%7Dt%5Cge0%5Cend%7Bcases%7D)
If we shift the argument by 1 and scale by -5, we have
![-5u(t-1)=\begin{cases}0&\text{for }t](https://tex.z-dn.net/?f=-5u%28t-1%29%3D%5Cbegin%7Bcases%7D0%26%5Ctext%7Bfor%20%7Dt%3C1%5C%5C-5%26%5Ctext%7Bfor%20%7Dt%5Cge1%5Cend%7Bcases%7D)
so if we subtract this from
![-5u(t)](https://tex.z-dn.net/?f=-5u%28t%29)
, we'll end up with
![-5u(t)+5u(t-1)=\begin{cases}0&\text{for }t](https://tex.z-dn.net/?f=-5u%28t%29%2B5u%28t-1%29%3D%5Cbegin%7Bcases%7D0%26%5Ctext%7Bfor%20%7Dt%3C0%5Ctext%7B%20and%20%7Dt%5Cge1%5C%5C-5%26%5Ctext%7Bfor%20%7D0%5Cle%20t%3C1%5Cend%7Bcases%7D)
For the next piece, we can add another scaled and shifted step like
![-6u(t-1)+6u(t-7)=\begin{cases}0&\text{for }t](https://tex.z-dn.net/?f=-6u%28t-1%29%2B6u%28t-7%29%3D%5Cbegin%7Bcases%7D0%26%5Ctext%7Bfor%20%7Dt%3C1%5Ctext%7B%20and%20%7Dt%5Cge7%5C%5C-6%26%5Ctext%7Bfor%20%7D1%5Cle%20t%3C7%5Cend%7Bcases%7D)
so that
![-5u(t)+5u(t-1)-6u(t-1)+6u(t-7)=\begin{cases}0&\text{for }t](https://tex.z-dn.net/?f=-5u%28t%29%2B5u%28t-1%29-6u%28t-1%29%2B6u%28t-7%29%3D%5Cbegin%7Bcases%7D0%26%5Ctext%7Bfor%20%7Dt%3C0%5Ctext%7B%20and%20%7Dt%5Cge7%5C%5C-5%26%5Ctext%7Bfor%20%7D0%5Cle%20t%3C1%5C%5C-6%26%5Ctext%7Bfor%20%7D1%5Cle%20t%3C7%5Cend%7Bcases%7D)
For the last piece, we add one more term:
![6u(t-7)=\begin{cases}0&\text{for }t](https://tex.z-dn.net/?f=6u%28t-7%29%3D%5Cbegin%7Bcases%7D0%26%5Ctext%7Bfor%20%7Dt%3C7%5C%5C6%26%5Ctext%7Bfor%20%7Dt%5Cge7%5Cend%7Bcases%7D)
and so putting everything together, we get
![f(t)](https://tex.z-dn.net/?f=f%28t%29)
:
![f(t)\equiv-5u(t)+5u(t-1)-6u(t-1)+6u(t-7)+6u(t-7)](https://tex.z-dn.net/?f=f%28t%29%5Cequiv-5u%28t%29%2B5u%28t-1%29-6u%28t-1%29%2B6u%28t-7%29%2B6u%28t-7%29)
Answer:
i think its a <em>knockoff</em>
Step-by-step explanation:
The area of the deck is :
A ( deck ) = 8 * 27.63 = 221.04 ft²
The area of the sheet of wood:
A ( sheet ) = 4 * 8 = 32 ft²
221.04 : 32 = 6.9 ≈ 7 ( we need a whole number )
Answer: He should buy 7 sheets of wood to cover the surface of the deck.
Answer:
Juan travel 31.4 feet farther than Fred in one rotation.
Step-by-step explanation:
In this problem we need to determine the change in linear position of Fred and Juan, whose formula is:
(Eq. 1)
Where:
- Radius, measured in feet.
- Angular arch, measured in radians.
- Change in linear position, measured in feet.
If both makes one rotation in the carousel, we obtain the change in linear position of each player:
Fred (
,
)
![s= 2\pi\cdot (7\,ft)](https://tex.z-dn.net/?f=s%3D%202%5Cpi%5Ccdot%20%287%5C%2Cft%29)
![s = (3.14)\cdot (7\,ft)](https://tex.z-dn.net/?f=s%20%3D%20%283.14%29%5Ccdot%20%287%5C%2Cft%29)
![s = 21.98\,ft](https://tex.z-dn.net/?f=s%20%3D%2021.98%5C%2Cft)
Juan (
,
)
![s= 2\pi\cdot (17\,ft)](https://tex.z-dn.net/?f=s%3D%202%5Cpi%5Ccdot%20%2817%5C%2Cft%29)
![s = (3.14)\cdot (17\,ft)](https://tex.z-dn.net/?f=s%20%3D%20%283.14%29%5Ccdot%20%2817%5C%2Cft%29)
![s = 53.38\,ft](https://tex.z-dn.net/?f=s%20%3D%2053.38%5C%2Cft)
And the difference between both travelled distances is:
![\Delta s = 53.38\,ft-21.98\,ft](https://tex.z-dn.net/?f=%5CDelta%20s%20%3D%2053.38%5C%2Cft-21.98%5C%2Cft)
![\Delta s = 31.4\,ft](https://tex.z-dn.net/?f=%5CDelta%20s%20%3D%2031.4%5C%2Cft)
Juan travel 31.4 feet farther than Fred in one rotation.