What is the midline of this function? y=     What is the amplitude of this function?  
   Define a function, g, to represent the behavior of the graphed function. g(a)= NO LINKS. ​

Answers:

There are infinitely many possible ways to answer the third part.

Explanation:

The midline is the horizontal line that goes through the center of the sine curve. Visually we can see that is y = 0. Another way we can see this is to note how y = 0 is the midpoint of y = 1.5 and y = -1.5, which are the max and min respectively.

The amplitude is the vertical distance from y = 0 to y = 1.5, and it's also the vertical distance from y = 0 to y = -1.5; in short, it's the vertical distance from center to either the peak or valley.

The general format of a sine curve is

y = A*sin(B(x-C)) + D

where,

In this case, we found so far that

The only thing we're missing is the value of C, which is the phase shift.

Note the point (pi/2, 1.5) is one of the max points on this curve. Also, recall that sin(x) maxes out at 1 when x = pi/2

This must mean that the stuff inside the sine, the B(x-C) portion, must be equal to pi/2 in order to lead sin(B(x-C)) = 1.

So,

B(x-C) = pi/2

0.5(pi/2-C) = pi/2

pi/4 - C/2 = pi/2

4*( pi/4 - C/2 ) = 4*(pi/2)

pi - 2C = 2pi

-2C = 2pi - pi

-2C = pi

C = pi/(-2)

C = -pi/2

This allows us to update the function to get g(x) = 1.5*sin(0.5(x+pi/2)) which is the same as g(x) = 1.5sin(0.5x + pi/4)

This is one possible answer because we could have infinitely many possible values for C, due to sin(x) = 1 having infinitely many solutions.

Also, you could use cosine instead of sine. Cosine is just a phase shifted version and of sine, and vice versa.

∠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.

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