Y= 3 cos(x)+2 I need help solving this

Use the form acos(bx−c)+dacos(bx-c)+d to find the variables used to find the amplitude, period, phase shift, and vertical shift.

a=3a=3

b=12b=12

c=0c=0

d=0d=0

Find the amplitude |a||a|.

Amplitude: 33

Find the period of 3cos(x2)3cos(x2).

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4π4π

Find the phase shift using the formula cbcb.

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Phase Shift: 00

Find the vertical shift dd.

Vertical Shift: 00

List the properties of the trigonometric function.

Amplitude: 33

Period: 4π4π

Phase Shift: 00 (00 to the right)

Vertical Shift: 00

Select a few points to graph.

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xf(x)03π02π−33π04π3xf(x)03π02π-33π04π3

The trig function can be graphed using the amplitude, period, phase shift, vertical shift, and the points.

Amplitude: 33

Period: 4π4π

Phase Shift: 00 (00 to the right)

Vertical Shift: 00

xf(x)03π02π−33π04π3xf(x)03π02π-33π04π3

to plot points   for x just start at the origin (middle) and go to the left however many times it say(if the number is a positive) if the number is a negative then go right however many times it says if it is 0 go nowhere

now keep you pencil there and go up however many times it says on y(if it is positive) if it is negative then go down however many times it says. if it says 0 go nowhere.

you have just plotted a point! now when you finish plotting all of them connect them with a ruler

plot each point as (-3, 7), (0,4), (3,1), (8,-4)   in the parenthesis on the left of the comma is x and the right is y.

∠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.