<span>30 hours
For this problem, going to assume that the actual flow rate for both pipes is constant for the entire duration of either filling or emptying the pool. The pipe to fill the pool I'll consider to have a value of 1/12 while the drain that empties the pool will have a value of 1/20. With those values, the equation that expresses how many hour it will take to fill the pool while the drain is open becomes:
X(1/12 - 1/20) = 1
Now solve for X
X(5/60 - 3/60) = 1
X(2/60) = 1
X(1/30) = 1
X/30 = 1
X = 30
To check the answer, let's see how much water would have been added over 30 hours.
30/12 = 2.5
So 2 and a half pools worth of water would have been added. Now how much would be removed?
30/20 = 1.5
And 1 and half pools worth would have been removed. So the amount left in the pool is
2.5 - 1.5 = 1
And that's exactly the amount needed.</span>
Answer:
2.9403x10^15
Step-by-step explanation:
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Answer:
This contradicts the Mean Value Theorem since there exists a c on (1, 7) such that f '(c) = f(7) − f(1) (7 − 1) , but f is not continuous at x = 3
Step-by-step explanation:
The given function is
When we differentiate this function with respect to x, we get;
We want to find all values of c in (1,7) such that f(7) − f(1) = f '(c)(7 − 1)
This implies that;
![c-3=\sqrt[3]{63.15789}](https://tex.z-dn.net/?f=c-3%3D%5Csqrt%5B3%5D%7B63.15789%7D)
![c=3+\sqrt[3]{63.15789}](https://tex.z-dn.net/?f=c%3D3%2B%5Csqrt%5B3%5D%7B63.15789%7D)
If this function satisfies the Mean Value Theorem, then f must be continuous on [1,7] and differentiable on (1,7).
But f is not continuous at x=3, hence this hypothesis of the Mean Value Theorem is contradicted.
