It is 1
or as an improper fraction it would be
<span><span>y = 2 + 2sec(2x)
The upper part of the range will be when the secant has the smallest
positive value up to infinity.
The smallest positive value of the secant is 1
So the minimum of the upper part of the range of
y = 2 + 2sec(2x) is 2 + 2(1) = 2 + 2 = 4
So the upper part of the range is [4, )
The lower part of the range will be from negative infinity
up to when the secant has the largest negative value.
The largest negative value of the secant is -1
So the maximum of the lower part of the range of
y = 2 + 2sec(2x) is 2 + 2(-1) = 2 - 2 = 0
So the lower part of the range is (, 0].
Therefore the range is (, 0] U [4, )
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The equation represented by Ms. Wilson's model is n² + 13n + 40 = (n + 8)(n + 5)
<h3>How to determine the equation of the model?</h3>
The partially completed model is given as:
| n
| n²
5 | 5n | 40
By dividing the rows and columns, the complete model is:
| n | 8
n | n² | 8n
5 | 5n | 40
Add the cells, and multiply the leading row and columns
n² + 8n + 5n + 40 = (n + 8)(n + 5)
This gives
n² + 13n + 40 = (n + 8)(n + 5)
Hence, the equation represented by Ms. Wilson's model is n² + 13n + 40 = (n + 8)(n + 5)
Read more about polynomials at:
brainly.com/question/4142886
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