The expression factorized completely is (h +2k)[(h+2k) + (2k-h)]
From the question,
We are to factorize the expression (h+2k)²+4k²-h² completely
The expression can be factorized as shown below
(h+2k)²+4k²-h² becomes
(h+2k)² + 2²k²-h²
(h+2k)² + (2k)²-h²
Using difference of two squares
The expression (2k)²-h² = (2k+h)(2k-h)
Then,
(h+2k)² + (2k)²-h² becomes
(h+2k)² + (2k +h)(2k-h)
This can be written as
(h+2k)² + (h +2k)(2k-h)
Now,
Factorizing, we get
(h +2k)[(h+2k) + (2k-h)]
Hence, the expression factorized completely is (h +2k)[(h+2k) + (2k-h)]
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Answer:

Step-by-step explanation:
i)
A 5 unit circle essay would be about whatever you would want it to be and this is no question this is cheating
4 is the nearest integer
as a decimal this number would be 3.5, which would round up to 4.
Plotting the data (attached photo) roughly shows that the data is skewed to the left. In other words, data is skewed negatively and that the long tail will be on the negative side of the peak.
In such a scenario, interquartile range is normally the best measure to compare variations of data.
Therefore, the last option is the best for the data provided.