Height of the water increasing is at rate of 
<h3>How to solve?</h3>
With related rates, we need a function to relate the 2 variables, in this case it is clearly volume and height. The formula is:
There is radius in the formula, but in this problem, radius is constant so it is not a variable. We can substitute the value in:
Since the rate in this problem is time related, we need to implicitly differentiate wrt (with respect to) time:
In the problem, we are given 
So we need to substitute this in:
Hence, Height of the water increasing is at rate of
<h3>Formula used: </h3>
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Answer:
x =−8
Step-by-step explanation:
To get the *percent increase* from week 1 to week 2, we calculate the change in distance from week 1 to week 2 (13.5 - 12.5 = 1 mile) over the week 1 distance (12.5 miles). Doing that, we find that Matthew increased his distance by
1/12.5 = 0.08, or 8%
We’re given that he’ll increase his distance by the same percentage from week 2 to 3, so to find his week 3 distance, we can find 8% of the week 2 distance and add that on. 8% of 13.5 miles is 0.08 x 13.5 = 1.08 miles, so by week 3, he’ll be running 13.5 + 1.08 = 14.58 miles.
Answer:
15.5
Step-by-step explanation:
Answer:
<h2 /><h2>The interquartile range (IQR) is the difference between the upper (Q3) and lower (Q1) quartiles, and describes the middle 50% of values when ordered from lowest to highest. The IQR is often seen as a better measure of spread than the range as it is not affected by outliers. Interquartile Range. 25% of values.</h2>
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<h2>here your answer </h2>
