Let point (a, b) be on a unit circle.
If (a, b) has a rotation of r degrees, then which of the following is not correct.
cos(-r) = -a
cos(r) = a
sin(-r) = -a
sin(r) = -a
sin(r) = b

The incorrect statements are:

Consider any angle in standard position together with a unit circle. The terminal side will intersect the circle at some point (x, y) .

Let the unit circle moves in the counterclockwise direction until the angle that is formed between our position, the origin, and the positive x-axis is equal to θ.

The y-coordinate of this point is always equal to the sine of the angle, and the x-coordinate of this point is always equal to the cosine of the angle.

That is  

sin θ= y

cosθ = x

Now, we have point (a, b) be on a unit circle.

If (a, b) has a rotation of r degrees.

Then

sin(-r) = -a,

sin(r) = -a is equal to the x-coordinate.

Therefore they are incorrect.

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(C) sin(-r) = -a

(D) sin(r) = -a

Explanation:

The unit circle definition allows us to extend the domain of sine and cosine to all real numbers. The process for determining the sine/cosine of any angle θ is as follows:

Starting from (1,0) move along the unit circle in the counterclockwise direction until the angle that is formed between your position, the origin, and the positive x-axis is equal to θ.

In the question, sin(-r) = -a and sin(r) = -a is equal to the x-coordinate, thus they are not correct.

An image is attached for reference.

the correct functions are

h(x) = 3x + 2  

g(x) = 2^x

step-by-step explanation:

i proceed to calculate the different values ​​of k (x)

case 1)  k(x) = g(x) ∘ h(x)

g(x) ∘ h(x)=g(h(x))

k(x) = [2]^[3x + > is the option  4. k(x) = 2^(3x + 2 )

case 2)  k(x) = g(x) + h(x)

k(x) = [2^x]+[3x + 2]=2^x+3x+> is the option  1.) k(x) = 2^x + 3x + 2

case 3)k(x) = h(x) ÷ g(x)

k(x) =  [3x + 2]/[2^x]=3x/(2^x)+2/(2^x)=3x*(2^-x)+2^(1-x)

is the option  3.)   k(x) = (3x)2^(-x) + 2^(-x + 1 )

case 4)k(x) = g(x) - h(x)

k(x) =  [2^x]-[3x + 2]=2^x-3x-> is the option  2.) k(x) = 2^x - 3x - 2 

case 5)  k(x) = g(x) × h(x)

k(x) = [2^x]*[3x + > is the option  5.) k(x) = 2^x(3x + 2)

case 6)k(x) = h(x) ∘ g(x)

h(x) ∘ g(x)=h(g(x))

k(x)=3*[2^x]+> is the option  6. k(x) = 3(2^x) +2

-7/8

step-by-step explanation:

instead of subtracting you add the -6/8 and -1/8