Which sum or difference identity would you use to verify that cos (180° - q) = -cos q? a.) sin (a -b) = sin a cos b – cos a sin b b.) cos (a -b) = cos a cos b – sin a sin b c.) cos (a -b) = cos a cosb + sin a sin b d.) sin (a + b) = sin a cos b + cos a sin b

The correct answer is Option C

Step-by-step explanation:

In this question first we use the sum and difference formula :

cos (a-b) = cos(a) cos(b) + sin (a) sin (b)

Let a = 180 and b = q, then

cos (180-q) = cos (180) cos (q) + sin (180) sin (q)

Now the sin of 180 degree is 0. Since 0 multiplying  with anything leaves 0, this makes sin (180) sin (q) = 0

cos (180-q) = cos (180) cos(q)

Now the cos of 180 degree is -1.

cos(180-q) = -1 cos(q), or cos (180-q) = -cos(q)

x= -3 and

step-by-step explanation:

to find the roots of a cubic, use division or factoring to find the factors then set equal to 0. as this is a difference of cubes, it factors into

where a is the cube root of .

here   this is 27 and the cube root of 27 is 3. so we write using the form:

.

set each factor to 0 and solve for x.

x-3=0

x=3

the quadratic expression does not factor and must be solved using the quadratic formula.

where a = 1, b=3, and c=9

this cubic has two complex roots.

maybe showing the graph would

step-by-step explanation:

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