Answer: Choice A

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Explanation:
Recall that
and
. The connection between tangent and cotangent is simply involving the reciprocal
From this, we can say,

In the second to last step, a pair of sine terms cancel. In the last step, a pair of cosine terms cancel.
All of this shows why
is identical to 
Therefore,
is an identity. In mathematics, an identity is when both sides are the same thing for any allowed input in the domain.
You can visually confirm that
is the same as
by graphing each function (use x instead of alpha). You should note that both curves use the exact same set of points to form them. In other words, one curve is perfectly on top of the other. I recommend making the curves different colors so you can distinguish them a bit better.
Answer:
Please check the explanation!
Step-by-step explanation:
Given the equation

As some of the absolute rules are:
NOW, let us solve!
Let us substitute all the table values
Putting x = -4
∵


So, when x = -4, then y = -7
Putting x = -3



when x = -3, then y = -5
Putting x = -2



when x = -2, then y = -3
Putting x = -1



when x = -1, then y = -1
Putting x = 0



when x = 0, then y = -3
Putting x = 1



when x = 1, then y = -5
The graph is also attached below.
Answer:
y =
x - 5
Step-by-step explanation:
Given f(0) = - 5 and f(4) = - 3 , then we have the coordinate points
(0, - 5) and (4, - 3)
The equation of a line in slope- intercept form is
y = mx + c ( m is the slope and c the y- intercept )
Calculate m using the slope formula
m =
with (x₁, y₁ ) = (0, - 5) and (x₂, y₂ ) = (4, - 3)
m =
=
= 
The line crosses the y- axis at (0, - 5) ⇒ c = - 5
y = f(x) =
x - 5 ← equation of line
I think I already answered it to you from the same question you asked earlier.
But anyways, the answer is in the image below and the steps are also in one of the images too.
Hope this helps! :)
y ≥ 1x+1
y ≥ x+1
Two points that are on the graph of y ≥ x+1 are (0,1)and (1,2).
Use the equation for finding the slope.

=
=1
The slope is 1.