I don’t know how to do this one.

Part (a)

The unit circle is centered at (0,0). This point is the origin.

The radius of the unit circle is 1

The term "unit" means "one", which describes the radius length.

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Part (b)

The equation of the unit circle is x^2+y^2 = 1

The general equation of any circle is (x-h)^2+(y-k)^2 = r^2

In this case, the unit circle has center (h,k) = (0,0) and radius r = 1.

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Part (c)

The following four points are on the unit circle

As shown below in the diagram. The idea is to plug each given coordinate into the equation from part (b), and solve for the missing variable.

For example, if we know x = 1, then...

x^2+y^2 = 1

1^2+y^2 = 1

1+y^2 = 1

y^2 = 1-1

y^2 = 0

y = sqrt(0)

y = 0

Meaning x = 1 leads to y = 0. So (x,y) = (1,0) is one point on the circle. The other parts are handled in a similar fashion.

step-by-step explanation:

always plug in x. in this case, x equals 4.

i'm going to first solve 4(2+3x)

1) 4(2+3(4)) use pemdas to solve.

i'm going to multiply whats in the parenthesis first, so 3 times 4=12.

2)4 (2+12) i'm then going to multiply 4 by both numbers in the parenthesis (because multiplying comes before adding)

4 times 2= 8.   4 times 12= 48

4) (8+48)= 56 (answer)

i'm then going to solve 8+12x and plug in 4 for x.

1) 8 +12(4)

multiplying comes before adding, so i'm gonna multiply 12 and 4.

12 times 4= 48.

2) 8+48=56

both equal the same!