The graph of a sinusoidal function intersects its midline at (0,2) and then has a minimum point at (3,-6) Write the formula of the function, where x is entered in radians.

 y = -8sin(1/3)π x + 2

Step-by-step explanation:

The equation of a cosine function is:    y = A cos(Bx - C) + D    where

Midline (D) = 2

(0, 2) is given as a point on the midline.  We only need the y-value.

Horizontal stretch (B) = (1/3)π

The min is located at (3,-6) --> x = 3  which is half of the Period.  Thus the period (length of one wave) is 3·2 = 6 units.

Phase Shift (C) = 0

The midline is on the y-axis so there is no horizontal shift.

Amplitude (A) = 7

The distance from the min to the midline is: A = -6 - 2 = -8

Equation

Input A = -8, B = (1/3)π, C = 0, and D = 2 into the sine equation.

            y = A sin(Bx - C) + D

            y = -8sin((1/3)π x - 0) + 2

            y = -8sin(1/3)π x + 2

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.