Please help; will give brainliest!
How would you graph this function?

Frequency = 1/6π

Amplitude = 2

Midline: y = 3

y-intercept: (0, 3)

The function is not a reflection of its parent function over the x-axis.

I missed class because of technical issues and I would love some help. I might have more questions so feel free to stick around. :)

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Step-by-step explanation:

So, when you see the function, you have to make it able to be graphed. So you would start with this equation: y = a*sin(bx-c)+d. Apply the function, you get: y = 2*sin(1/3*x-0)+3. Simplify that down, and the resulting function is: y = 2*sin(x/3)+3

∠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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step-by-step explanation: