First tell me what elapsed time is
so, is a semi-circle, half a circle, recall a circle has a total of 360°, so half of that will be 180°.
the diameter of that circle is 10, so its radius is half that, or 5.
![\bf \textit{arc's length}\\\\ s=\cfrac{\theta \pi r}{180}~~ \begin{cases} r=radius\\ \theta =angle~in\\ \qquad degrees\\[-0.5em] \hrulefill\\ \theta =180\\ r=5 \end{cases}\implies s=\cfrac{(180)(\pi )(5)}{180}\implies s=5\pi \stackrel{\pi =3.14}{\implies s=15.7}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Barc%27s%20length%7D%5C%5C%5C%5C%20s%3D%5Ccfrac%7B%5Ctheta%20%5Cpi%20r%7D%7B180%7D~~%20%5Cbegin%7Bcases%7D%20r%3Dradius%5C%5C%20%5Ctheta%20%3Dangle~in%5C%5C%20%5Cqquad%20degrees%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20%5Ctheta%20%3D180%5C%5C%20r%3D5%20%5Cend%7Bcases%7D%5Cimplies%20s%3D%5Ccfrac%7B%28180%29%28%5Cpi%20%29%285%29%7D%7B180%7D%5Cimplies%20s%3D5%5Cpi%20%5Cstackrel%7B%5Cpi%20%3D3.14%7D%7B%5Cimplies%20s%3D15.7%7D)
0.21 is greater than 0.12 because it in the 20s
Answer:
Machine's useful number of years = 9 years
Step-by-step explanation:
Using the straight line method, depreciation is calculated as the difference between the cost of the equipment minus the salvage value, all divided by the number of useful years.
Yearly Depreciation
= (Cost - Salvage value) ÷ (Number of useful years)
Yearly depreciation = P20,000
Cost = P200,000
Salvage Value = P20,000
Number of useful years = n
20000 = (200000 - 20000) ÷ n
20000 = (180000/n)
n = (180000/20000) = 9 years
Hope this Helps!!!
Answer:
$46.00
Step-by-step explanation:
$46.00 - $20 + $5 = $21
