Is it ever possible that cos(a−b)=cos ⁡a − cos ⁡b?

yes

Step-by-step explanation:

I'm going to give you a slightly different answer, but it's going to make sense :-)

First, let's review what "sin" and "cos" really mean. They are functions that take as an input an angle, which we call theta. They output the base (cos) and height (sin) of a triangle which as a hypotenuse of length 1.    

Now, let's pick some examples. If we happen to set theta to 45 degress, you will get a triangle that looks like this:

In this case, both sin(theta) and cos(theta) are the same number, the square root of 1/2. So cos(theta) + cos (theta) is 2 times the square tool of 1/2.

Now imagine that we now want to find cos (theta + theta). Remember that theta was 45 degrees, so this will be cos (45 + 45), or cos (90).

But remember that cos is the base of a triangle where theta is the angle with the base. Well, that's not a triangle at all, is it? It's just a vertical line. In fact, cos(90) will be zero.

the smallest possible measurement of the width is 14 feet

step-by-step explanation:

let's assume length of parallelogram is 'l'

width of parallelogram is 'w'

the perimeter of a parallelogram must be no less than 40 feet

so, we get

we are given

length of parallelogram is 6 feet

so,

l=6

we can plug it

now, we can solve for w

subtract both sides by 12

divide both sides by 2

and we get

so,

the smallest possible measurement of the width is 14 feet

answer: (c) m∠qpo + (2x + 16)° = 180°

step-by-step explanation:

the opposite angles of a quadrilateral are supplementary.

so, ∠o + ∠q = 180°   and   ∠p + ∠r = 180°

since ∠r = 2x + 16°, we can use substitution as follows:

∠p + ∠r = 180°

∠p + (2x + 16)° = 180°