Answer:
Step-by-step explanation:
Solution for Solve for x. Assume that lines which appear tangent are tangent. 5) 13x + 2 T. U V 48° S. 130° M. ... A: Given: fx,y,z=x2y3+z4 and x=s2, y=st2 and z=s2t. a) Calculate the primary derivatives ∂f∂x, ∂f∂y, ∂f
L = rate of the large hose
S = rate of the small hose
we know the sum of their rates deliver the swimming pool filled up in 35 minutes, thus L + S = 35.
now, the large hose can do it in 60 minutes, that means that in 1 minute, the large hose has only done 1/60 th of the work.
the small hose can do the whole thing in S minutes, that means in 1 minute it has only done 1/S th of the whole work.
and since both of them working together can do it in 35 minutes, then in 1 minute they both have done 1/35 th of the whole job.
![\bf \stackrel{\textit{large's rate in 1 minute}}{\cfrac{1}{L}}+\stackrel{\textit{small's rate in 1 minute}}{\cfrac{1}{S}}=\stackrel{\textit{done in 1 minute}}{\cfrac{1}{35}} \\\\\\ \cfrac{1}{60}+\cfrac{1}{S}=\cfrac{1}{35}\implies \cfrac{S+60}{60S}=\cfrac{1}{35}\implies 35S+2100=60S\implies 2100=25S \\\\\\ \cfrac{2100}{25}=S\implies 84=S](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Ctextit%7Blarge%27s%20rate%20in%201%20minute%7D%7D%7B%5Ccfrac%7B1%7D%7BL%7D%7D%2B%5Cstackrel%7B%5Ctextit%7Bsmall%27s%20rate%20in%201%20minute%7D%7D%7B%5Ccfrac%7B1%7D%7BS%7D%7D%3D%5Cstackrel%7B%5Ctextit%7Bdone%20in%201%20minute%7D%7D%7B%5Ccfrac%7B1%7D%7B35%7D%7D%20%5C%5C%5C%5C%5C%5C%20%5Ccfrac%7B1%7D%7B60%7D%2B%5Ccfrac%7B1%7D%7BS%7D%3D%5Ccfrac%7B1%7D%7B35%7D%5Cimplies%20%5Ccfrac%7BS%2B60%7D%7B60S%7D%3D%5Ccfrac%7B1%7D%7B35%7D%5Cimplies%2035S%2B2100%3D60S%5Cimplies%202100%3D25S%20%5C%5C%5C%5C%5C%5C%20%5Ccfrac%7B2100%7D%7B25%7D%3DS%5Cimplies%2084%3DS)
Answer: 300
Explanation:
The upper quartile, or quartile 3, is the median of the upper half of the set. In this case, the upper quartile is 300.
The answer is b in case you were wondering so yea 11:522