<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, )
</span>
</span>
The slope m of a line
through points

and

is given by :<span>

Thus, the slope of the line passing through points </span><span>(−1, 7) and (2, 10) is
</span>

<span>
The equation of a line with slope m passing through a point P(a, b) is given by
(y-b)=m(x-a).
We can consider any of the points (-1, 7), or (2, 10). Let's choose (2, 10):
y-10=1(x-2)
y-10=x-2
y-x=-2+10
y-x=8
Answer: </span><span>C. −x+y=8</span>
Answer:
g= (9,8) f= (-5,1)
Step-by-step explanation:
you're welcome =)
So you can choose any number from 11 to 19.
I'll choose 11 for now.
Then you write, 11, the number, and then spell it out, eleven.
To show how many tens and ones, you can make a place value chart. You would make two boxes, the first one for tens and the other for ones.
You would put a 1 in the tens place and a 1 in the ones place.
I hope this helps!