I don’t know how to do this one.
2 answers:
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
- (1,0)
- (0,1)
- (-1,0)
- (0,-1)
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.
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Answer:
Option b
Step-by-step explanation:
A function is even if and only if f(-x) = f(x).
This means that there is the same value of y for and for
Where is any point belonging to the function .
Therefore you must select a graph whose ordered pairs are of the form and
Therefore the correct answer is option b. Assume that each box in the graph equals 1 unit. Then, for x = 2 y = -3 and for x = -2 y = -3. Then it is shown that the function is even.