It will increase..........................
Answer:
the state of giving off light or glow.
Answer:
- The formula its
![f(t) \ = \ - \ 352 \ \frac{\$ }{years} \ t \ + \ \$ \ 2816](https://tex.z-dn.net/?f=f%28t%29%20%5C%20%3D%20%5C%20-%20%5C%20352%20%5C%20%5Cfrac%7B%5C%24%20%7D%7Byears%7D%20%5C%20t%20%5C%20%2B%20%5C%20%5C%24%20%5C%202816%20)
- After 5 years, the computer value its $ 1056
Explanation:
<h3>
Obtaining the formula</h3>
We wish to find a formula that
- Starts at 2816.
![f(0 \ years) \ = \ \$ \ 2816](https://tex.z-dn.net/?f=f%280%20%5C%20years%29%20%5C%20%3D%20%5C%20%5C%24%20%5C%202816)
- Reach 0 at 8 years.
![f( 8 \ years) \ = \ \$ \ 0](https://tex.z-dn.net/?f=f%28%208%20%5C%20years%29%20%5C%20%3D%20%5C%20%5C%24%20%5C%200)
- Depreciates at a constant rate. m
We can cover all this requisites with a straight-line equation. (an straigh-line its the only curve that has a constant rate of change) :
,
where m its the slope of the line and b give the place where the line intercepts the <em>y</em> axis.
So, we can use this formula with the data from our problem. For the first condition:
![f ( 0 \ years ) = m \ (0 \ years) + b = \$ \ 2816](https://tex.z-dn.net/?f=f%20%28%200%20%5C%20years%20%29%20%3D%20m%20%5C%20%280%20%5C%20years%29%20%2B%20b%20%3D%20%5C%24%20%5C%202816)
![b = \$ \ 2816](https://tex.z-dn.net/?f=%20b%20%3D%20%5C%24%20%5C%202816)
So, b = $ 2816.
Now, for the second condition:
![f ( 8 \ years ) = m \ (8 \ years) + \$ \ 2816 = \$ \ 0](https://tex.z-dn.net/?f=f%20%28%208%20%5C%20years%20%29%20%3D%20m%20%5C%20%288%20%5C%20years%29%20%2B%20%5C%24%20%5C%202816%20%3D%20%5C%24%20%5C%200)
![m \ (8 \ years) = \ - \$ \ 2816](https://tex.z-dn.net/?f=%20m%20%5C%20%288%20%5C%20years%29%20%3D%20%5C%20-%20%5C%24%20%5C%202816)
![m = \frac{\ - \$ \ 2816}{8 \ years}](https://tex.z-dn.net/?f=%20m%20%3D%20%5Cfrac%7B%5C%20-%20%5C%24%20%5C%202816%7D%7B8%20%5C%20years%7D)
![m = \frac{\ - \$ \ 2816}{8 \ years}](https://tex.z-dn.net/?f=%20m%20%3D%20%5Cfrac%7B%5C%20-%20%5C%24%20%5C%202816%7D%7B8%20%5C%20years%7D)
![m = \ - \ 352 \frac{\$ }{years}](https://tex.z-dn.net/?f=%20m%20%3D%20%5C%20-%20%5C%20352%20%5Cfrac%7B%5C%24%20%7D%7Byears%7D)
So, our formula, finally, its:
![f(t) \ = \ - \ 352 \ \frac{\$ }{years} \ t \ + \ \$ \ 2816](https://tex.z-dn.net/?f=f%28t%29%20%5C%20%3D%20%5C%20-%20%5C%20352%20%5C%20%5Cfrac%7B%5C%24%20%7D%7Byears%7D%20%5C%20t%20%5C%20%2B%20%5C%20%5C%24%20%5C%202816%20)
<h3>After 5 years</h3>
Now, we just use <em>t = 5 years</em> in our formula
![f(5 \ years) \ = \ - \ 352 \ \frac{\$ }{years} \ 5 \ years \ + \ \$ \ 2816](https://tex.z-dn.net/?f=f%285%20%5C%20years%29%20%5C%20%3D%20%5C%20-%20%5C%20352%20%5C%20%5Cfrac%7B%5C%24%20%7D%7Byears%7D%20%5C%205%20%5C%20years%20%5C%20%2B%20%5C%20%5C%24%20%5C%202816%20)
![f(5 \ years) \ = \ - \$ \ 1760 + \ \$ \ 2816](https://tex.z-dn.net/?f=f%285%20%5C%20years%29%20%5C%20%3D%20%5C%20-%20%5C%24%20%5C%201760%20%2B%20%5C%20%5C%24%20%5C%202816%20)
![f(5 \ years) \ = $ \ 1056](https://tex.z-dn.net/?f=f%285%20%5C%20years%29%20%5C%20%3D%20%24%20%5C%201056%20)
Answer:
83.3 kHz
Explanation:
The frequency of a waveform is equal to the reciprocal of its period:
![f=\frac{1}{T}](https://tex.z-dn.net/?f=f%3D%5Cfrac%7B1%7D%7BT%7D)
where
f is the frequency
T is the period
In this problem, we have
![T=12 \mu s=12\cdot 10^{-6} s](https://tex.z-dn.net/?f=T%3D12%20%5Cmu%20s%3D12%5Ccdot%2010%5E%7B-6%7D%20s)
so, the frequency of the waveform is
![f=\frac{1}{12 \cdot 10^{-6} s}=8.33\cdot 10^4 Hz](https://tex.z-dn.net/?f=f%3D%5Cfrac%7B1%7D%7B12%20%5Ccdot%2010%5E%7B-6%7D%20s%7D%3D8.33%5Ccdot%2010%5E4%20Hz)
And by converting into kiloHertz,
![f=8.33\cdot 10^4 Hz=83.3 kHz](https://tex.z-dn.net/?f=f%3D8.33%5Ccdot%2010%5E4%20Hz%3D83.3%20kHz)