What function best represents a sine function with an amplitude of 4, a period of pi/6 , and a midline at y = −2?

a. f(x) = −2 sin(x − π/6 ) + 4

b. f(x) = 4 sin 12x − 2

c. f(x) = 4 sin(x − π/6 ) − 2

d. f(x) = −2 sin 12x + 4

B). f(x) = 4 sin 12x − 2.

Step-by-step explanation:

Given : A sine function with an amplitude of 4, a period of pi/6 , and a midline at y = −2.

To find : What function best represents a sine function.

Solution : We have  amplitude of 4, a period of pi/6 , and a midline at y = −2.

Standard form of sin function : y = a sin[ b(x-h)] + k.

Where, a = amplitude

= time period.

b =

b = 12x

k = vertical displacement. , h = 0

Then Substitute the values  a =4 , b = 12x k = -2 in standard form

y = 4 sin 12x - 2.

Therefore, B). f(x) = 4 sin 12x − 2.

-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3

168

step-by-step explanation:

i am pretty sure that the answer is 168