<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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You subtract and find x=10
Let the three odd integers be x - 2, x and x + 2
x - 2 + x = 3(x + 2) + 7
2x - 2 = 3x + 6 + 7
2x - 2 = 3x + 13
3x - 2x = -2 - 13
x = -15
The three consecutive odd integers are -17, -15 and -13.
Here’s the answer w explanation :)
