recall your d = rt, distance = rate * time.
let's say we have two trains, A and B, A is going at 85 mph and B at 65 mph.
they are 210 miles apart and moving toward each other, at some point they will meet, when that happens, the faster train A has covered say d miles, and the slower B has covered then the slack from 210 and d, namely 210 - d.
When both trains meet, A has covered more miles than B because A is faster, however the time both have been moving, is the same, say t hours.
![\bf \begin{array}{lcccl} &\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\ \cline{2-4}&\\ \textit{Train A}&d&85&t\\ \textit{Train B}&210-d&65&t \end{array}\\\\ \dotfill\\\\ \begin{cases} \boxed{d}=85t\\ 210-d=65t\\[-0.5em] \hrulefill\\ 210-\boxed{85t}=65t \end{cases} \\\\\\ 210=150t\implies \cfrac{210}{150}=t\implies \cfrac{7}{5}=t\implies \stackrel{\textit{one hour and 24 minutes}}{1\frac{2}{5}=t}](https://tex.z-dn.net/?f=%20%5Cbf%20%5Cbegin%7Barray%7D%7Blcccl%7D%20%26%5Cstackrel%7Bmiles%7D%7Bdistance%7D%26%5Cstackrel%7Bmph%7D%7Brate%7D%26%5Cstackrel%7Bhours%7D%7Btime%7D%5C%5C%20%5Ccline%7B2-4%7D%26%5C%5C%20%5Ctextit%7BTrain%20A%7D%26d%2685%26t%5C%5C%20%5Ctextit%7BTrain%20B%7D%26210-d%2665%26t%20%5Cend%7Barray%7D%5C%5C%5C%5C%20%5Cdotfill%5C%5C%5C%5C%20%5Cbegin%7Bcases%7D%20%5Cboxed%7Bd%7D%3D85t%5C%5C%20210-d%3D65t%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20210-%5Cboxed%7B85t%7D%3D65t%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20210%3D150t%5Cimplies%20%5Ccfrac%7B210%7D%7B150%7D%3Dt%5Cimplies%20%5Ccfrac%7B7%7D%7B5%7D%3Dt%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bone%20hour%20and%2024%20minutes%7D%7D%7B1%5Cfrac%7B2%7D%7B5%7D%3Dt%7D%20)
The numerator in a fraction represents the number of pieces selected. If the numerator is larger than the denominator, the number is larger than one
<span>The denominator in a fraction represents the total number of equal size pieces that make up a whole
</span>
Answer:
$10.00 per hour
Step-by-step explanation:
Overhead application rate which is also known as overhead absorption rate on the basis on labor hours is the total budgeted overhead for 2019 which is $900,000 divided by the expected production of 90,000 labor hours for the year.
overhead application rate=$900,000/90,000=$10 per hour
This implies that for every one hour worked overhead cost of $10 would be added to the other costs incurred.
The correct option then is the third option of $10.00 per hour
Answer:
1
Step-by-step explanation:
Because it y is 0 then we can find the X intercepts. so there will be one X intercept.
